存个BM板子

B Eddy Walker 2

#include <bits/stdc++.h>
#define eps 1e-8
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lson l,mid,rt<<1
#define rson mid+1,r,(rt<<1)+1
#define CLR(x,y) memset((x),y,sizeof(x))
#define fuck(x) cerr << #x << "=" << x << endl
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int seed = 131;
const int maxn = 1e5 + 5;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
#define SZ(x) ((ll)(x).size())
typedef vector<ll> VI;
typedef pair<ll, ll> PII;
const ll mod = 1000000007;
ll powmod(ll a, ll b) {
    ll res = 1;
    a %= mod;
    assert(b >= 0);
    for (; b; b >>= 1) {
        if (b & 1)res = res * a % mod;
        a = a * a % mod;
    }
    return res;
}
// head

ll _;
namespace linear_seq {
    const ll N = 10010;
    ll res[N], base[N], _c[N], _md[N];

    vector<ll> Md;
    void mul(ll *a, ll *b, ll k) {
        rep(i, 0, k + k) _c[i] = 0;
        rep(i, 0, k) if (a[i]) rep(j, 0, k) _c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;
        for (ll i = k + k - 1; i >= k; i--) if (_c[i])
                rep(j, 0, SZ(Md)) _c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
        rep(i, 0, k) a[i] = _c[i];
    }
    ll solve(ll n, VI a, VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
    //        printf("%d\n",SZ(b));
        ll ans = 0, pnt = 0;
        ll k = SZ(a);
        assert(SZ(a) == SZ(b));
        rep(i, 0, k) _md[k - 1 - i] = -a[i];
        _md[k] = 1;
        Md.clear();
        rep(i, 0, k) if (_md[i] != 0) Md.push_back(i);
        rep(i, 0, k) res[i] = base[i] = 0;
        res[0] = 1;
        while ((1ll << pnt) <= n) pnt++;
        for (ll p = pnt; p >= 0; p--) {
            mul(res, res, k);
            if ((n >> p) & 1) {
                for (ll i = k - 1; i >= 0; i--) res[i + 1] = res[i];
                res[0] = 0;
                rep(j, 0, SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;
            }
        }
        rep(i, 0, k) ans = (ans + res[i] * b[i]) % mod;
        if (ans < 0) ans += mod;
        return ans;
    }
    VI BM(VI s) {
        VI C(1, 1), B(1, 1);
        ll L = 0, m = 1, b = 1;
        rep(n, 0, SZ(s)) {
            ll d = 0;
            rep(i, 0, L + 1) d = (d + (ll)C[i] * s[n - i]) % mod;
            if (d == 0) ++m;
            else if (2 * L <= n) {
                VI T = C;
                ll c = mod - d * powmod(b, mod - 2) % mod;
                while (SZ(C) < SZ(B) + m) C.pb(0);
                rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
                L = n + 1 - L;
                B = T;
                b = d;
                m = 1;
            } else {
                ll c = mod - d * powmod(b, mod - 2) % mod;
                while (SZ(C) < SZ(B) + m) C.pb(0);
                rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
                ++m;
            }
        }
        return C;
    }
    ll gao(VI a, ll n) {
        VI c = BM(a);
        c.erase(c.begin());
        rep(i, 0, SZ(c)) c[i] = (mod - c[i]) % mod;
        return solve(n, c, VI(a.begin(), a.begin() + SZ(c)));
    }
};
int T;
ll k, n;
ll dp[5005];
int main() {
    scanf("%d", &T);
    while (T--) {
        CLR(dp, 0);
        scanf("%lld%lld", &k, &n);
        if (n == 0) {
            printf("1\n");
            continue;
        }
        if (n == -1) {
            printf("%lld\n", 2LL * powmod(k + 1, mod - 2) % mod);
            continue;
        }
        vector<ll>v;
        dp[0] = 1;
        v.push_back(1);
        for (int i = 1; i <= k; i++) {
            for (int j = 0; j < i; j++) {
                dp[i] = (dp[i] + dp[j]) % mod;
            }
            dp[i] = dp[i] * powmod(k, mod - 2) % mod;
            v.push_back(dp[i]);
        }
        for (int i = k + 1; i <= 2 * k; i++) {
            for (int j = 1; j <= min(int(k), i); j++) {
                dp[i] = (dp[i] + dp[i - j]) % mod;
            }
            dp[i] = dp[i] * powmod(k, mod - 2) % mod;
            v.push_back(dp[i]);
        }
        printf("%lld\n", linear_seq::gao(v, n));

    }
    return 0;
}