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:
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#define MAXN 3010
#define MOD 1000000007
#define INV 500000004
inline void trim(int &x) { (x >= MOD) && (x -= MOD); }
inline void up(int &x, int y) { trim(x += y); }
int n, q, b = 1, ans = 0, a[MAXN], f[MAXN][MAXN];
int main() {
	scanf("%d%d", &n, &q);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			f[i][j] = a[i] > a[j];
		}
	}
	for (int x, y; q; q--) {
		scanf("%d%d", &x, &y);
		for (int i = 1; i <= n; i++) {
			if (i != x && i != y) {
				f[i][x] = f[i][y] = 1LL * INV * (f[i][x] + f[i][y]) % MOD;
				f[x][i] = f[y][i] = 1LL * INV * (f[x][i] + f[y][i]) % MOD;
			}
		}
		f[x][y] = f[y][x] = 1LL * INV * (f[x][y] + f[y][x]) % MOD;
		trim(b <<= 1);
	}
	for (int i = 1; i < n; i++) {
		for (int j = i + 1; j <= n; j++) {
			up(ans, f[i][j]);
		}
	}
	printf("%lld\n", 1LL * ans * b % MOD);
	return 0;
}