Problem: <image>

To monitor cryptocurrency exchange rates trader William invented a wonderful device consisting of n lights arranged in a row. The device functions in the following way:

Initially, all lights on William's device are turned off. At the beginning of a new iteration the device randomly, with a uniform distribution, picks a light that is turned off and turns it on, telling William which cryptocurrency he should invest in. After this iteration if any k consecutive lights contain more than one turned on light, then the device finishes working.

William doesn't like uncertainty, so he wants you to calculate the expected value of the number of lights that are turned on in the device after it finishes working.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.

The only line for each test case contains two integers n and k (2 ≤ k ≤ n ≤ 10^5), which are the total number of lights and the length of subsegment of lights that are being checked, respectively.

Output

For each test case print the answer, modulo 10^9+7. 

Formally, let M = 10^9+7. It can be shown that the answer can be expressed as an irreducible fraction p/q, where p and q are integers and q not ≡ 0 \pmod{M}. Output the integer equal to p ⋅ q^{-1} mod M. In other words, output such an integer x that 0 ≤ x < M and x ⋅ q ≡ p \pmod{M}.

Example

Input


3
3 2
15 2
40 15


Output


333333338
141946947
329622137

Note

Explanation of the first sample test case:

Let's write out all possible sequences of light toggles, which will make the device complete its operation:

  1. (1, 2) — 2 lights are turned on 
  2. (1, 3, 2) — 3 lights are turned on 
  3. (2, 1) — 2 lights are turned on 
  4. (2, 3) — 2 lights are turned on 
  5. (3, 2) — 2 lights are turned on 
  6. (3, 1, 2) — 3 lights are turned on 



Then the final expected value will be equal to 2/6 + 3/6 + 2/6 + 2/6 + 2/6 + 3/6 = 14/6 = 7/3. 

Then the required output will be 333333338, since 333333338 ⋅ 3 ≡ 7 \pmod{10^9+7}.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

#define f first
#define s second
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define ll long long
#define ull unsigned long long
#define db double
#define ld long double
#define pii pair<int, int>
#define pil pair<int, ll>
#define pli pair<ll, int>
#define pll pair<ll, ll>
#define px pair<db, db>
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define input(a) for (auto& el : a) cin >> el;
#define output(a) cout << '!'; for (auto& el : a) cout << el << ' '; cout << '\n'; cout.flush();
#define sz(el) int(el.size())
#define sq(a) ((a) * (a))
#define xll __int128
#define xmin(a,b) a = min(a, b)
#define xmax(a,b) a = max(a, b)
#define sun(a) sort(all(a)); a.resize(unique(all(a)) - a.begin());

using namespace std;

template <class T1, class T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& p) {
    return (out << p.f << ' ' << p.s << endl);
}

const int INF = 1000000017;
const ll LLINF = 1000000000000000017LL;
const ld inf = 1.0 / 0.0;
const ld eps = 1e-10;
const int MOD = 1e9 + 7;
const int MOD2 = 1e9 + 9;

mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count());

//uniform_int_distribution<int> uid(l, r); [l, r]
//int x = uid(rng);

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> tret;

/// *find_by_order - val by index, order_of_key - # less than x
*/

/*struct Action {
    int t, i, j;

    Action(int t, int i, int j) : t(t), i(i), j(j) {}
};*/

/// Author: @mathalex
namespace modint {
    static const int MODx = 1e9 + 7;

    struct mint {
        int a = 0;

        mint() {}

        mint(int a) : a(a) {
            a = (a % MODx + MODx) % MODx;
        }

        mint operator-() const {
            if (a) {
                return -a + MODx;
            } else {
                return 0;
            }
        }

        mint operator+(const mint& b) const {
            int res = a + b.a;
            if (res >= MODx) {
                res -= MODx;
            }
            return res;
        }

        void operator+=(const mint& b) {
            a += b.a;
            if (a >= MODx) {
                a -= MODx;
            }
        }

        mint operator-(const mint& b) const {
            int res = a - b.a;
            if (res < 0) {
                res += MODx;
            }
            return res;
        }

        void operator-=(const mint& b) {
            a -= b.a;
            if (a < 0) {
                a += MODx;
            }
        }

        mint operator*(const mint& b) const {
            return (ll) a * b.a % MODx;
        }

        void operator*=(const mint& b) {
            a = (ll) a * b.a % MODx;
        }

        int get() const {
            return a;
        }
    };

    mint fpow(mint a, int p) {
        mint x = 1;
        while (p) {
            if (p & 1) {
                x *= a;
            }
            a *= a;
            p >>= 1;
        }
        return x;
    }

    istream& operator>>(istream& in, mint& a) {
        return (in >> a.a);
    }

    ostream& operator<<(ostream& out, const mint& a) {
        return (out << a.a);
    }
}

using namespace modint;

const int N = 1e5 + 7;

mint f[N], rf[N];

mint c(ll n, ll k) {
    if (n < k || n < 0) {
        return 0;
    }
    return f[n] * rf[k] * rf[n - k];
}

mint rc(ll n, ll k) {
    return rf[n] * f[k] * f[n - k];
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    f[0] = rf[0] = 1;
    for (int i = 1; i < N; ++i) {
        f[i] = f[i - 1] * i;
        rf[i] = rf[i - 1] * fpow(i, MODx - 2);
    }

    int NT;
    cin >> NT;
    while (NT--) {
        int n, k;
        cin >> n >> k;
        vector<mint> p(n + 2, 1);
        for (int i = 2; i <= n + 1; ++i) {
            p[i] = c(n - 1ll * (i - 2) * (k - 1), i - 1) * rc(n, i - 1);
        }
        //output(p);

        mint res = 0;
        for (int i = 2; i <= n; ++i) {
            res += (p[i] - p[i + 1]) * i;
        }
        cout << res << '\n';
    }
    return 0;
}