#include<bits/stdc++.h>
using namespace std;
vector <int> v[200005];
int n,i,c0,cost[200005],q,j,x,t,y,d[800005],dp[200005],b[200005];
int query(int k,int t,int w,int x,int y)
{
	if ((t>y) || (x>w))
	  return 2000000000;
	if ((x<=t) && (w<=y))
	  return d[k];
	int mid=(t+w)/2;
	return min(query(k*2,t,mid,x,y),query(k*2+1,mid+1,w,x,y));
}
void xiugai(int k,int t,int w,int x,int y)
{
	if ((t>x) || (w<x))
	  return;
	if (t==w)
	{
		d[k]=y;
		return;
    }
    int mid=(t+w)/2;
    xiugai(k*2,t,mid,x,y);
    xiugai(k*2+1,mid+1,w,x,y);
    d[k]=min(d[k*2],d[k*2+1]);
}
int main()
{
	scanf("%d",&n);
	for (i=1;i<=n;i++)
	  {
	  	scanf("%d",&b[i]);
	  	if (b[i]==0)
	  	  c0++;
	  	if (b[i]==0)
	  	  cost[i]=-1;
	  	else
	  	  cost[i]=1;
	  }
	scanf("%d",&q);
	for (i=1;i<=q;i++)
	{
		scanf("%d%d",&x,&y);
		v[x].push_back(y);
	}
	memset(d,127,sizeof(d));
	memset(dp,127,sizeof(dp));
	dp[0]=0;
	for (i=1;i<=n;i++)
    {
    	for (j=0;j<v[i].size();j++)
    	{
    		x=v[i][j];
    		t=min(dp[i-1],query(1,1,n,max(i-1,1),x));
    		if (t<dp[x])
    		{
    			dp[x]=t;
    		    xiugai(1,1,n,x,t);
			}
		}
	dp[i]=min(dp[i],dp[i-1]+cost[i]);
	}
    printf("%d",dp[n]+c0);
    return 0;
}
### Prompt

Please formulate a CPP solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
vector <int> v[200005];
int n,i,c0,cost[200005],q,j,x,t,y,d[800005],dp[200005],b[200005];
int query(int k,int t,int w,int x,int y)
{
	if ((t>y) || (x>w))
	  return 2000000000;
	if ((x<=t) && (w<=y))
	  return d[k];
	int mid=(t+w)/2;
	return min(query(k*2,t,mid,x,y),query(k*2+1,mid+1,w,x,y));
}
void xiugai(int k,int t,int w,int x,int y)
{
	if ((t>x) || (w<x))
	  return;
	if (t==w)
	{
		d[k]=y;
		return;
    }
    int mid=(t+w)/2;
    xiugai(k*2,t,mid,x,y);
    xiugai(k*2+1,mid+1,w,x,y);
    d[k]=min(d[k*2],d[k*2+1]);
}
int main()
{
	scanf("%d",&n);
	for (i=1;i<=n;i++)
	  {
	  	scanf("%d",&b[i]);
	  	if (b[i]==0)
	  	  c0++;
	  	if (b[i]==0)
	  	  cost[i]=-1;
	  	else
	  	  cost[i]=1;
	  }
	scanf("%d",&q);
	for (i=1;i<=q;i++)
	{
		scanf("%d%d",&x,&y);
		v[x].push_back(y);
	}
	memset(d,127,sizeof(d));
	memset(dp,127,sizeof(dp));
	dp[0]=0;
	for (i=1;i<=n;i++)
    {
    	for (j=0;j<v[i].size();j++)
    	{
    		x=v[i][j];
    		t=min(dp[i-1],query(1,1,n,max(i-1,1),x));
    		if (t<dp[x])
    		{
    			dp[x]=t;
    		    xiugai(1,1,n,x,t);
			}
		}
	dp[i]=min(dp[i],dp[i-1]+cost[i]);
	}
    printf("%d",dp[n]+c0);
    return 0;
}
```