Problem: You are given an integer sequence of length N: A_1,A_2,...,A_N. Let us perform Q operations in order. The i-th operation is described by two integers X_i and Y_i. In this operation, we will choose one of the following two actions and perform it:

* Swap the values of A_{X_i} and A_{Y_i}
* Do nothing



There are 2^Q ways to perform these operations. Find the sum of the inversion numbers of the final sequences for all of these ways to perform operations, modulo 10^9+7.

Here, the inversion number of a sequence P_1,P_2,...,P_M is the number of pairs of integers (i,j) such that 1\leq i < j\leq M and P_i > P_j.

Constraints

* 1 \leq N \leq 3000
* 0 \leq Q \leq 3000
* 0 \leq A_i \leq 10^9(1\leq i\leq N)
* 1 \leq X_i,Y_i \leq N(1\leq i\leq Q)
* X_i\neq Y_i(1\leq i\leq Q)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
A_1
:
A_N
X_1 Y_1
:
X_Q Y_Q


Output

Print the sum of the inversion numbers of the final sequences, modulo 10^9+7.

Examples

Input

3 2
1
2
3
1 2
1 3


Output

6


Input

5 3
3
2
3
1
4
1 5
2 3
4 2


Output

36


Input

9 5
3
1
4
1
5
9
2
6
5
3 5
8 9
7 9
3 2
3 8


Output

425

Solution (in CPP):

Solution:
#if 0
date +%Y.%m.%d

题解先咕着。
#endif
#include <cstdio>
#include <algorithm>
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long ll;

struct {
	inline operator int () { int x; return scanf("%d", &x), x; }
} read;

const int maxn = 3030, mod = 1000000007;
int a[maxn];
ll f[maxn][maxn];

int main () {
	int n = read, q = read;
	for (int i = 1; i <= n; i ++) a[i] = read;
	for (int i = 1; i <= n; i ++)
		for (int j = 1; j <= n; j ++)
			f[i][j] = a[i] > a[j];

	ll coe = 1, i2 = (mod + 1) >> 1;
	while (q --) {
		int x = read, y = read;
		for (int i = 1; i <= n; i ++)
			if (i != x and i != y)
				f[i][y] = f[i][x] = i2 * (f[i][x] + f[i][y]) % mod;
		for (int i = 1; i <= n; i ++)
			if (i != x and i != y)
				f[y][i] = f[x][i] = i2 * (f[x][i] + f[y][i]) % mod;
		f[y][x] = f[x][y] = i2 * (f[x][y] + f[y][x]) % mod;
		(coe <<= 1) %= mod;
	}

	/* for (int i = 1; i <= n; i ++) { */
	/* 	for (int j = 1; j <= n; j ++) */
	/* 		debug(" %lld", f[i][j]); */
	/* 	debug("\n"); */
	/* } */

	ll sum = 0;
	for (int i = 1; i <= n; i ++)
		for (int j = i + 1; j <= n; j ++)
			sum += f[i][j];
	sum %= mod;
	printf("%lld\n", sum * coe % mod);
}