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 maxn = 3005, mod = 1000000007;

int n, m, a[maxn];
ll f[maxn][maxn];

ll qpow(ll x, int y)
{
	ll ret = 1;
	while(y)
	{
		if(y&1)
			ret = ret * x % mod;
		x = x * x % mod;
		y >>= 1;
	}
	return ret;
}

int main()
{
	scanf("%d%d", &n, &m);
	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]);
	int u, v, inv2 = qpow(2, mod - 2);
	for(int i = 1; i <= m; ++i)
	{
		scanf("%d%d", &u, &v);
		if(u > v)	swap(u, v);
		f[u][v] = f[v][u] = inv2 * (f[u][v] + f[v][u]) % mod;
		for(int j = 1; j <= n; ++j)
			if(j != u && j != v)
			{
				f[j][u] = f[j][v] = inv2 * (f[j][u] + f[j][v]) % mod;
				f[u][j] = f[v][j] = inv2 * (f[u][j] + f[v][j]) % mod;
			}
	}
	ll ans = 0;
	for(int i = 1; i <= n; ++i)
		for(int j = i + 1; j <= n; ++j)
			ans = (ans + f[i][j]) % mod;
	printf("%lld", ans * qpow(2, m) % mod);
	return 0;
}