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 <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

const int N=2e5+5,M=N*4;

typedef long long LL;

int n,q,a[N],f[N],s[N],ans,T[M];

vector <int> h[N];

vector <int> ::iterator it;

char c;

int read()
{
	int x=0,sig=1;
	for (c=getchar();c<'0' || c>'9';c=getchar()) if (c=='-') sig=-1;
	for (;c>='0' && c<='9';c=getchar()) x=x*10+c-48;
	return x*sig;
}

void Ins(int l,int r,int g,int v,int x)
{
	if (l==r)
	{
		T[x]=min(T[x],v); return;
	}
	int mid=l+r>>1;
	if (g<=mid) Ins(l,mid,g,v,x<<1);else Ins(mid+1,r,g,v,x<<1|1);
	T[x]=min(T[x<<1],T[x<<1|1]);
}

int Query(int l,int r,int a,int b,int x)
{
	if (l==a && r==b) return T[x];
	int mid=l+r>>1;
	if (b<=mid) return Query(l,mid,a,b,x<<1);
	if (a>mid) return Query(mid+1,r,a,b,x<<1|1);
	return min(Query(l,mid,a,mid,x<<1),Query(mid+1,r,mid+1,b,x<<1|1));
}

int main()
{
	n=read();
	for (int i=1;i<=n;i++)
	{
		a[i]=read(); s[i]=s[i-1]+(a[i]^1);
	}
	q=read();
	for (int l,r;q--;)
	{
		l=read(); r=read(); h[l].push_back(r);
	}
	memset(f,127,sizeof(f));
	memset(T,127,sizeof(T));
	f[0]=ans=0;
	for (int i=1;i<=n;i++)
	{
		for (it=h[i].begin();it!=h[i].end();it++)
		{
			int j=s[*it]+min(Query(1,n,i,*it,1),ans-s[i-1]);
			Ins(1,n,*it,j-s[*it],1);
		}
		ans=min(ans+a[i],Query(1,n,i,i,1)+s[i]);
	}
	printf("%d\n",ans);
	return 0;
}