【DP+线段树转移矩阵】CF Good Bye 2016 E. New Year and Old Subsequence

http://codeforces.com/contest/750/problem/E

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

input

Copy

8 3
20166766
1 8
1 7
2 8

output

Copy

4
3
-1

input

Copy

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15

output

Copy

-1
2
1
-1
-1

input

Copy

4 2
1234
2 4
1 2

output

Copy

-1
-1

Note

In the first sample:

• In the first query, ugliness("20166766") = 4 because all four sixes must be removed.
• In the second query, ugliness("2016676") = 3 because all three sixes must be removed.
• In the third query, ugliness("0166766") =  - 1 because it's impossible to remove some digits to get a nice string.

In the second sample:

• In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice.
• In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170".
• 给你一个字符串，Q次询问，询问区间中一定要有子序列2017，一定没有子序列2016的最小删除个数是多少，若是不行输出-1
• val[i][j]:表示该区间有’2017’中的[i,j)子序列而没有[i,j]子序列，并且不会出现’2016’，所需要删除的最少字符数
• 比如val[0][2]表示该区间能够出现’20’但不能出现’201’需要删除的最少字符数。又如arr[1][1]表示不会出现’1’。
• 需要注意的是’6’不能出现在[3][3]（即’1’之后，’7’之前）和[4][4]（即’7’之后）。

#include <bits/stdc++.h>
using namespace std;
const int INF=1e9;
const int maxn=2e5+10;
struct node
{
    int val[5][5];
    node() //新建初始化
    {
        for(int i=0;i<5;i++)
            for(int j=0;j<5;j++)
                val[i][j]=INF;
    }

    node operator + (node x) //重载运算符 矩阵相加
    {
        node ans;
        for(int i=0;i<5;i++)
            for(int j=0;j<5;j++)
                for(int k=0;k<5;k++)
                    ans.val[i][j]=min(ans.val[i][j],val[i][k]+x.val[k][j]);
        return ans;
    }
}tree[maxn*4];

char ch[10]={'2','0','1','7','6'};
char s[maxn];

node pushup(node &a,node &b)
{
    node res;
    for(int i=0;i<5;i++)
        for(int j=i;j<5;j++)
            for(int k=i;k<=j;k++)
                res.val[i][j]=min(res.val[i][j],a.val[i][k]+b.val[k][j]);
    return res;
}

int Find(char c)
{
    for(int i=0;i<5;i++)
    {
        if(ch[i]==c)
            return i;
    }
    return -1;
}

void build(int l,int r,int rt)
{
    if(l==r)
    {
        int t=Find(s[l]);
        for(int i=0;i<5;i++)
        {
            tree[rt].val[i][i]=0;
        }
        if(t!=-1&&t<4) //是2017中的
        {
            tree[rt].val[t][t+1]=0;
            tree[rt].val[t][t]=1;
        }
        else if(t==4) //有6
        {
            tree[rt].val[3][3]=tree[rt].val[4][4]=1;
        }
        return;
    }
    int mid=(l+r)/2;
    build(l,mid,2*rt);
    build(mid+1,r,2*rt+1);
    tree[rt]=pushup(tree[2*rt],tree[2*rt+1]);
}

node query(int l,int r,int rt,int L,int R)
{
    if(L<=l&&R>=r)
    {
        return tree[rt];
    }
    int mid=(l+r)/2;
    if(mid<L) return query(mid+1,r,2*rt+1,L,R);
    if(mid>=R) return query(l,mid,2*rt,L,R);
    return query(l,mid,2*rt,L,R)+query(mid+1,r,2*rt+1,L,R); //注意一定要左区间在前，右区间在后，不然答案会不对
}

int main()
{
    int n,m,l,r;
    scanf("%d%d%s",&n,&m,s+1);
    build(1,n,1);
    while(m--)
    {
        node ans;
        scanf("%d%d",&l,&r);
        ans=query(1,n,1,l,r);
        if(ans.val[0][4]==INF)
            printf("-1\n");
        else
            printf("%d\n",ans.val[0][4]);
    }
    return 0;
}