#include<bits/stdc++.h>
using namespace std;
const int maxn = 3005, mod = 1e9 + 7;
int N, M, Right[maxn], s[maxn];
long long dp[maxn][maxn];
char str[maxn];
inline int S(int L, int R){return (s[R] - s[L - 1]);}

int main()
{
	scanf("%d%d", &N, &M);
	scanf("%s", str + 1);
	for(int i = 1; i <= N + 1; ++i) s[i] = s[i - 1] + (str[i] == '1');
	for(int i = 1; i <= N + 1; ++i) Right[i] = i;
	for(int L, R, i = 1; i <= M; ++i){
		scanf("%d%d", &L, &R);
		Right[L] = max(Right[L], R);
	}
	for(int i = 1; i <= N; ++i) Right[i] = max(Right[i], Right[i - 1]);
	dp[1][s[Right[1]]] = 1;
	for(int i = 1; i <= N; ++i)
		for(int j = 0; j <= N; ++j)
			if(dp[i][j]){
				int lin = S(Right[i] + 1, Right[i + 1]);
				int one = j, zero = Right[i] - (i - 1) - j;
				if(one) dp[i + 1][j - 1 + lin] = (dp[i + 1][j - 1 + lin] + dp[i][j]) % mod;  
				if(zero) dp[i + 1][j + lin] = (dp[i + 1][j + lin] + dp[i][j]) % mod;
			}
	cout << dp[N + 1][0];
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

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


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
const int maxn = 3005, mod = 1e9 + 7;
int N, M, Right[maxn], s[maxn];
long long dp[maxn][maxn];
char str[maxn];
inline int S(int L, int R){return (s[R] - s[L - 1]);}

int main()
{
	scanf("%d%d", &N, &M);
	scanf("%s", str + 1);
	for(int i = 1; i <= N + 1; ++i) s[i] = s[i - 1] + (str[i] == '1');
	for(int i = 1; i <= N + 1; ++i) Right[i] = i;
	for(int L, R, i = 1; i <= M; ++i){
		scanf("%d%d", &L, &R);
		Right[L] = max(Right[L], R);
	}
	for(int i = 1; i <= N; ++i) Right[i] = max(Right[i], Right[i - 1]);
	dp[1][s[Right[1]]] = 1;
	for(int i = 1; i <= N; ++i)
		for(int j = 0; j <= N; ++j)
			if(dp[i][j]){
				int lin = S(Right[i] + 1, Right[i + 1]);
				int one = j, zero = Right[i] - (i - 1) - j;
				if(one) dp[i + 1][j - 1 + lin] = (dp[i + 1][j - 1 + lin] + dp[i][j]) % mod;  
				if(zero) dp[i + 1][j + lin] = (dp[i + 1][j + lin] + dp[i][j]) % mod;
			}
	cout << dp[N + 1][0];
	return 0;
}
```