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>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int MOD = 1e9 + 7;
const int INV2 = (MOD + 1) / 2;

int dp[3005][3005];
int n, p[3005];
PII pinv[3005];
int m, u[3005], v[3005];
int cdp[3005];

int main()
{
	scanf("%d%d", &n, &m);
	rep1(i, n) scanf("%d", &p[i]);
	rep(i, m) scanf("%d%d", &u[i], &v[i]);
	
	rep1(i, n) dp[i][i] = 1;
	rep(i, m) rep1(j, n) {
		int cur = 1LL * (dp[j][u[i]] + dp[j][v[i]]) * INV2 % MOD;
		dp[j][u[i]] = dp[j][v[i]] = cur;
	}
	
	rep1(i, n) pinv[i] = MP(p[i], i);
	sort(pinv + 1, pinv + n + 1);
	int ans = 0;
	rep1(i, n) {
		int tot = 0;
		for(int j = n; j >= 1; j --) {
			ans = (ans + 1LL * (tot + 1LL * cdp[j] * INV2 % MOD) * dp[pinv[i].second][j]) % MOD;
			tot = (tot + cdp[j]) % MOD;
		}
		if(i == n || pinv[i].first != pinv[i + 1].first)
		for(int i0 = i; i0 >= 0 && pinv[i0].first == pinv[i].first; i0 --)
		rep1(j, n) cdp[j] = (cdp[j] + dp[pinv[i0].second][j]) % MOD;
	}
	
	rep(i, m) ans = ans * 2 % MOD;
	printf("%d\n", ans);
	return 0;
}