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<cstdio>
#include<cstring>
#include<algorithm>
#include<assert.h>
#define fo(i,a,b) for(int i=(a);i<=(b);++i)
#define fd(i,b,a) for(int i=(b);i>=(a);--i)
#define mset(a,x) memset(a,x,sizeof(a))
#define mcpy(a,b) memcpy(a,b,sizeof(b))
using namespace std;
typedef long long ll;
int read(){int n=0,p=1;char ch;for(ch=getchar();ch<'0' || ch>'9';ch=getchar())if(ch=='-') p=-1;for(;'0'<=ch && ch<='9';ch=getchar()) n=n*10+ch-'0';return n*p;}
const int N=3005,mo=1e9+7,iv2=5e8+4;
int n,a[N];
ll f[N][N],tf[N][N];
void inc(ll &x,ll y){x=(x+y)%mo;}
int main()
{
	n=read();int Q=read();
	fo(i,1,n) a[i]=read();
	fo(i,1,n)
		fo(j,1,n) f[i][j]=(a[i]>a[j]);
	fo(k,1,Q)
	{
		int x=read(),y=read();
		fo(i,1,n)
			if(i!=x && i!=y)
				f[i][x]=f[i][y]=(f[i][x]+f[i][y])*iv2%mo,f[x][i]=f[y][i]=(f[x][i]+f[y][i])*iv2%mo;
		f[x][y]=f[y][x]=(f[x][y]+f[y][x])*iv2%mo;
	}
	ll ans=0;
	fo(i,1,n-1)
		fo(j,i+1,n) inc(ans,f[i][j]);
	fo(i,1,Q) ans=ans*2%mo;
	printf("%lld\n",ans);
	return 0;
}