#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>

#define pb push_back
#define mp make_pair

using namespace std;

typedef unsigned int u32;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;

const lod pi=acos(-1);
const int oo=(1<<30)-1;
const LL OO=1e18;

const int N=2e5+100;

int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}

int mi[N*4];
#define lc i<<1
#define rc lc|1
inline void push(int i,int l,int r,int L,int R,int t) {
	if (L<=l&&r<=R) mi[i]=min(mi[i],t);
	else {
		int mid=(l+r)/2;
		if (L<=mid) push(lc,l,mid,L,R,t);
		if (mid<R) push(rc,mid+1,r,L,R,t);
	}
}
inline int query(int i,int l,int r,int k) {
	if (l==r) return mi[i];
	int mid=(l+r)/2;
	return min(mi[i],k<=mid?query(lc,l,mid,k):query(rc,mid+1,r,k));
}

int pre[N],sum[N],f[N];
#define l first
#define r second
PR p[N];
int main()
{
	int n,m,i,cur=0,w,l,r;
	for (i=1,n=gi();i<=n;i++) sum[i]=sum[i-1]+gi();
	int ans=sum[n];
	for (i=1,m=gi();i<=m;i++) p[i].r=gi(),p[i].l=gi();
	sort(p+1,p+1+m);
	for (i=n*4;i;i--) mi[i]=oo;
	for (i=1;i<=m;i++) {
		l=p[i].r,r=p[i].l;
		while (cur<r)
			pre[cur+1]=pre[cur],cur++;
		w=(r-l+1)-(sum[r]-sum[l-1])+sum[l-1]+pre[l-1];
		w=min(w,r-sum[r]+query(1,1,n,l));
		pre[r]=min(pre[r],w-sum[r]);
		push(1,1,n,l,r,w+sum[r]-r);
		ans=min(ans,w+sum[n]-sum[r]);
	}
	cout<<ans<<endl;
	return 0;
}

### Prompt

Your task is to create 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<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>

#define pb push_back
#define mp make_pair

using namespace std;

typedef unsigned int u32;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;

const lod pi=acos(-1);
const int oo=(1<<30)-1;
const LL OO=1e18;

const int N=2e5+100;

int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}

int mi[N*4];
#define lc i<<1
#define rc lc|1
inline void push(int i,int l,int r,int L,int R,int t) {
	if (L<=l&&r<=R) mi[i]=min(mi[i],t);
	else {
		int mid=(l+r)/2;
		if (L<=mid) push(lc,l,mid,L,R,t);
		if (mid<R) push(rc,mid+1,r,L,R,t);
	}
}
inline int query(int i,int l,int r,int k) {
	if (l==r) return mi[i];
	int mid=(l+r)/2;
	return min(mi[i],k<=mid?query(lc,l,mid,k):query(rc,mid+1,r,k));
}

int pre[N],sum[N],f[N];
#define l first
#define r second
PR p[N];
int main()
{
	int n,m,i,cur=0,w,l,r;
	for (i=1,n=gi();i<=n;i++) sum[i]=sum[i-1]+gi();
	int ans=sum[n];
	for (i=1,m=gi();i<=m;i++) p[i].r=gi(),p[i].l=gi();
	sort(p+1,p+1+m);
	for (i=n*4;i;i--) mi[i]=oo;
	for (i=1;i<=m;i++) {
		l=p[i].r,r=p[i].l;
		while (cur<r)
			pre[cur+1]=pre[cur],cur++;
		w=(r-l+1)-(sum[r]-sum[l-1])+sum[l-1]+pre[l-1];
		w=min(w,r-sum[r]+query(1,1,n,l));
		pre[r]=min(pre[r],w-sum[r]);
		push(1,1,n,l,r,w+sum[r]-r);
		ans=min(ans,w+sum[n]-sum[r]);
	}
	cout<<ans<<endl;
	return 0;
}

```