#include <iostream>
using namespace std;
#pragma warning (disable: 4996)

int mod = 1000000007;
int prev_[100009], dp[100009], dpr[100009];
int mulr[100009];

long long Q;
long long N[1 << 18], C[1 << 18], Base[1 << 18];
long long Answer[1 << 18];

void init(int pos) {
	int lims = (pos - 2) * (pos - 1) / 2;
	int lim2 = max(1, lims);
	if (pos == 1) {
		for (int i = lims; i <= 100000; i++) dp[i] = 1;
	}
	else {
		for (int i = lims; i <= 100000; i++) prev_[i] = dp[i];
		for (int i = lims; i <= 100000; i++) dp[i] = 0;
		for (int i = lims; i <= 100000 - pos; i++) dp[i + pos - 1] = prev_[i];
		for (int i = lims; i <= 100000 - pos; i++) {
			dp[i + pos] += dp[i];
			if (dp[i + pos] >= mod) dp[i + pos] -= mod;
		}
		for (int i = 0; i < lims; i++) dp[i] = 0;
	}
	for (int i = lims; i <= 100000; i++) dpr[i] = dp[i];
	for (int i = lim2; i <= 100000; i++) {
		dpr[i] += dpr[i - 1];
		if (dpr[i] >= mod) dpr[i] -= mod;
	}
	for (int i = lims; i <= 100000; i++) mulr[i] = (1LL * i * dp[i]) % mod;
	for (int i = lim2; i <= 100000; i++) {
		mulr[i] += mulr[i - 1];
		if (mulr[i] >= mod) mulr[i] -= mod;
	}
}

int main() {
	scanf("%lld", &Q);
	for (int i = 1; i <= Q; i++) scanf("%lld%lld", &N[i], &C[i]);
	for (int i = 1; i <= Q; i++) Base[i] = N[i] / C[i];
	for (int i = 1; i <= Q; i++) {
		if (Base[i] == 0) Answer[i] = N[i];
	}
	
	for (int gen = 1; gen <= 450; gen++) {
		init(gen); int lims = gen * (gen - 1) / 2;
		for (int i = 1; i <= Q; i++) {
			if (Base[i] <= lims) continue;
			int cl = Base[i] - 2 * gen + 1;
			int cr = Base[i] - gen; if (cr < 0) continue;
			
			// 1 個目の計算
			long long V1 = 0;
			if (cr >= 0) V1 = dpr[cr];
			if (cl >= 1) V1 -= dpr[cl - 1];
			if (V1 < 0) V1 += mod;
			
			long long S1 = 0;
			if (cr >= 0) S1 = mulr[cr];
			if (cl >= 1) S1 -= mulr[cl - 1];
			if (S1 < 0) S1 += mod;
			
			// 2 個目の計算
			long long V2 = 0;
			if (cl >= 1) V2 = dp[cl - 1];
			
			long long S2 = 0;
			if (cl >= 1) S2 = 1LL * (cl - 1 + gen) * dp[cl - 1] % mod;
			
			// 答えに加算
			Answer[i] += (1LL * V1 * N[i] - 1LL * S1 * C[i] + 1LL * mod * mod);
			Answer[i] += (1LL * V2 * N[i] - 1LL * S2 * C[i] + 1LL * mod * mod);
			Answer[i] %= mod; //cout << gen << ": " << V1 << " " << S1 << " " << V2 << " " << S2 << " " << Answer[i] << endl;
		}
	}
	
	for (int i = 1; i <= Q; i++) printf("%lld\n", Answer[i]);
	return 0;
}
### Prompt

Your challenge is to write a cpp solution to the following problem:

Eli- 1 started a part-time job handing out leaflets for N seconds. Eli- 1 wants to hand out as many leaflets as possible with her special ability, Cloning. Eli- gen can perform two kinds of actions below.

* Clone herself and generate Eli- (gen + 1) . (one Eli- gen (cloning) and one Eli- (gen + 1) (cloned) exist as a result of Eli-gen 's cloning.) This action takes gen \times C ( C is a coefficient related to cloning. ) seconds.
* Hand out one leaflet. This action takes one second regardress of the generation ( =gen ).



They can not hand out leaflets while cloning. Given N and C , find the maximum number of leaflets Eli- 1 and her clones can hand out in total modulo 1000000007 (= 10^9 + 7).

Constraints

* 1 \leq Q \leq 100000 = 10^5
* 1 \leq N_q \leq 100000 = 10^5
* 1 \leq C_q \leq 20000 = 2 \times 10^4

Input

The input is given from Standard Input in the following format:


Q
N_1 C_1
:
N_Q C_Q


The input consists of multiple test cases. On line 1 , Q that represents the number of test cases is given. Each test case is given on the next Q lines. For the test case q ( 1 \leq q \leq Q ) , N_q and C_q are given separated by a single space. N_q and C_q represent the working time and the coefficient related to cloning for test case q respectively.

Output

For each test case, Print the maximum number of leaflets Eli- 1 and her clones can hand out modulo 1000000007 ( = 10^9 + 7 ).

Partial Scores

30 points will be awarded for passing the test set satisfying the condition: Q = 1 .

Another 270 points will be awarded for passing the test set without addtional constraints and you can get 300 points in total.

Examples

Input

2
20 8
20 12


Output

24
20


Input

1
20 3


Output

67


Input

1
200 1


Output

148322100

### Response

```cpp
#include <iostream>
using namespace std;
#pragma warning (disable: 4996)

int mod = 1000000007;
int prev_[100009], dp[100009], dpr[100009];
int mulr[100009];

long long Q;
long long N[1 << 18], C[1 << 18], Base[1 << 18];
long long Answer[1 << 18];

void init(int pos) {
	int lims = (pos - 2) * (pos - 1) / 2;
	int lim2 = max(1, lims);
	if (pos == 1) {
		for (int i = lims; i <= 100000; i++) dp[i] = 1;
	}
	else {
		for (int i = lims; i <= 100000; i++) prev_[i] = dp[i];
		for (int i = lims; i <= 100000; i++) dp[i] = 0;
		for (int i = lims; i <= 100000 - pos; i++) dp[i + pos - 1] = prev_[i];
		for (int i = lims; i <= 100000 - pos; i++) {
			dp[i + pos] += dp[i];
			if (dp[i + pos] >= mod) dp[i + pos] -= mod;
		}
		for (int i = 0; i < lims; i++) dp[i] = 0;
	}
	for (int i = lims; i <= 100000; i++) dpr[i] = dp[i];
	for (int i = lim2; i <= 100000; i++) {
		dpr[i] += dpr[i - 1];
		if (dpr[i] >= mod) dpr[i] -= mod;
	}
	for (int i = lims; i <= 100000; i++) mulr[i] = (1LL * i * dp[i]) % mod;
	for (int i = lim2; i <= 100000; i++) {
		mulr[i] += mulr[i - 1];
		if (mulr[i] >= mod) mulr[i] -= mod;
	}
}

int main() {
	scanf("%lld", &Q);
	for (int i = 1; i <= Q; i++) scanf("%lld%lld", &N[i], &C[i]);
	for (int i = 1; i <= Q; i++) Base[i] = N[i] / C[i];
	for (int i = 1; i <= Q; i++) {
		if (Base[i] == 0) Answer[i] = N[i];
	}
	
	for (int gen = 1; gen <= 450; gen++) {
		init(gen); int lims = gen * (gen - 1) / 2;
		for (int i = 1; i <= Q; i++) {
			if (Base[i] <= lims) continue;
			int cl = Base[i] - 2 * gen + 1;
			int cr = Base[i] - gen; if (cr < 0) continue;
			
			// 1 個目の計算
			long long V1 = 0;
			if (cr >= 0) V1 = dpr[cr];
			if (cl >= 1) V1 -= dpr[cl - 1];
			if (V1 < 0) V1 += mod;
			
			long long S1 = 0;
			if (cr >= 0) S1 = mulr[cr];
			if (cl >= 1) S1 -= mulr[cl - 1];
			if (S1 < 0) S1 += mod;
			
			// 2 個目の計算
			long long V2 = 0;
			if (cl >= 1) V2 = dp[cl - 1];
			
			long long S2 = 0;
			if (cl >= 1) S2 = 1LL * (cl - 1 + gen) * dp[cl - 1] % mod;
			
			// 答えに加算
			Answer[i] += (1LL * V1 * N[i] - 1LL * S1 * C[i] + 1LL * mod * mod);
			Answer[i] += (1LL * V2 * N[i] - 1LL * S2 * C[i] + 1LL * mod * mod);
			Answer[i] %= mod; //cout << gen << ": " << V1 << " " << S1 << " " << V2 << " " << S2 << " " << Answer[i] << endl;
		}
	}
	
	for (int i = 1; i <= Q; i++) printf("%lld\n", Answer[i]);
	return 0;
}
```