#include <bits/stdc++.h>

#define MAXN (3010)
#define mod (1000000007)

int n, m;
int sumOne[MAXN], dp[MAXN][MAXN], right[MAXN];  //dp[i][j] means [0, i) has j ones
char str[MAXN];

int main(){
	scanf("%d%d", &n, &m);
	scanf("%s", str);
	for(int i = 0; i < n; ++ i){
		sumOne[i] = (i ? sumOne[i - 1] : 0) + str[i] - '0';
		right[i] = i;
	}
	for(int i = 1; i <= m; ++ i){
		int l, r;
		scanf("%d%d", &l, &r);
		-- l; -- r;
		for(int j = l; j <= r; ++ j){
			right[j] = std::max(right[j], r);
		}
	}
	dp[0][0] = 1;
	for(int i = 0; i < n; ++ i){
		for(int usedOne = 0; usedOne < i + 1; ++ usedOne){
			int r = right[i];
			int restOne = sumOne[r] - usedOne;
			if(restOne < 0) continue;
			
			if(restOne > 0){   //fill str[i] with '1'
				(dp[i + 1][usedOne + 1] += dp[i][usedOne]) %= mod;
			}
			
			if(r - i + 1 != restOne){  //fill str[i] with '0'
			    (dp[i + 1][usedOne] += dp[i][usedOne]) %= mod;
			}
		}
	}
	printf("%d\n", dp[n][sumOne[n - 1]]);
	return 0;
}
### Prompt

Create a solution in CPP for the following problem:

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>

#define MAXN (3010)
#define mod (1000000007)

int n, m;
int sumOne[MAXN], dp[MAXN][MAXN], right[MAXN];  //dp[i][j] means [0, i) has j ones
char str[MAXN];

int main(){
	scanf("%d%d", &n, &m);
	scanf("%s", str);
	for(int i = 0; i < n; ++ i){
		sumOne[i] = (i ? sumOne[i - 1] : 0) + str[i] - '0';
		right[i] = i;
	}
	for(int i = 1; i <= m; ++ i){
		int l, r;
		scanf("%d%d", &l, &r);
		-- l; -- r;
		for(int j = l; j <= r; ++ j){
			right[j] = std::max(right[j], r);
		}
	}
	dp[0][0] = 1;
	for(int i = 0; i < n; ++ i){
		for(int usedOne = 0; usedOne < i + 1; ++ usedOne){
			int r = right[i];
			int restOne = sumOne[r] - usedOne;
			if(restOne < 0) continue;
			
			if(restOne > 0){   //fill str[i] with '1'
				(dp[i + 1][usedOne + 1] += dp[i][usedOne]) %= mod;
			}
			
			if(r - i + 1 != restOne){  //fill str[i] with '0'
			    (dp[i + 1][usedOne] += dp[i][usedOne]) %= mod;
			}
		}
	}
	printf("%d\n", dp[n][sumOne[n - 1]]);
	return 0;
}
```