Problem: You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
int n,q,a[202020],b[202020],c[202020],t[202020*4];
vector<int>g[202020];
void gx(int l,int r,int nm)
{
	l+=202020;
	r+=202020;
	while(l<=r)
	{
		t[l]=min(t[l],nm);
		t[r]=min(t[r],nm);
		l=(l+1)/2;
		r=(r-1)/2;
	}
}
int gt(int x)
{
	int ans=1e9;
	x+=202020;
	while(x)
	{
		ans=min(ans,t[x]);
		x/=2;
	}
	return ans;
}
int main()
{
	for(int i=0;i<202020*3;i++)
	{
		t[i]=1e9;
	}
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
		b[i]=1-a[i]+b[i-1];
	}
	cin>>q;
	while(q--)
	{
		int a,b;
		cin>>a>>b;
		g[b].push_back(a);
	}
	for(int i=1;i<=n;i++)
	{
		gx(i-1,i-1,c[i-1]-b[i-1]);
		c[i]=c[i-1]+a[i];
		for(int j=0;j<g[i].size();j++)
		{
			int l=g[i][j],r=i;
			gx(l,r,gt(l-1));
		}
		c[i]=min(c[i],gt(i)+b[i]);
	}
	cout<<c[n]<<endl;
	return 0;
}