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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int maxn = 3005;
const int inv2 = (mod + 1) / 2;
int a[maxn];
int dp[maxn][maxn];
void add(int& a, int b) {
	a += b;
	if(a >= mod) a -= mod;
}
int main() {
	int n, q;
	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++) dp[i][j] = (a[i] < a[j]);
	int tot = 1;
	while(q--) {
		tot = tot * 2 % mod;
		int x, y;
		scanf("%d%d", &x, &y);
		for(int i = 1; i <= n; i++) 
			if(i != x && i != y) {
				dp[i][x] = dp[i][y] = (ll)(dp[i][x] + dp[i][y]) * inv2 % mod;
				dp[x][i] = dp[y][i] = (ll)(dp[x][i] + dp[y][i]) * inv2 % mod;
			}
		dp[x][y] = dp[y][x] = (ll)(dp[x][y] + dp[y][x]) * inv2 % mod;
	}
	int ans = 0;
	for(int i = 1; i <= n; i++) 
		for(int j = 1; j < i; j++) add(ans, dp[i][j]);
	ans = (ll)ans * tot % mod;
	printf("%d\n", ans);
	return 0;
}