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 N 3005
using namespace std;
typedef long long ll;
const ll mod = 1000000007;
int n,q,a[N],x,y;
int now = 1,las = 0;
ll f[N][N],inv2;
ll qp(ll a,ll b) { ll ret = 1; for(;b;b>>=1,a = a * a % mod) if(b & 1) ret = ret * a % mod; return ret; }


int main()
{
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;++i) scanf("%d",a + i);
	inv2 = qp(2,mod - 2);
	ll qpn = qp(2,q);
	for(int i=1;i<=n;++i)
	  for(int j=1;j<=n;++j)
	    if(a[i] > a[j]) f[i][j] = 1;
	while(q--)
	{
		scanf("%d%d",&x,&y);
		swap(now,las);
		for(int i=1;i<=n;++i) 
		{
			if(i == x || i == y) continue;
			f[i][x] = f[i][y] = (f[i][x] + f[i][y]) % mod * inv2 % mod;
			f[x][i] = f[y][i] = (f[x][i] + f[y][i]) % mod * inv2 % mod;
		}
		f[y][x] = f[x][y] = (f[x][y] + f[y][x]) % mod * inv2 % mod;
	}
	ll ans = 0;
	for(int i=1;i<=n;++i)
	  for(int j=i + 1;j<=n;++j)
	     ans = (ans + f[i][j] * qpn % mod) % mod;
	cout<<ans<<endl;
	return 0;
}