/* --- author: dxm --- */
#include<bits/stdc++.h>
using namespace std;

#define INF 1e9
#define Inf 1e18
#define rep(i,n) for(int (i)=0;(i)<n;i++)
#define REP(i,n) for(int (i)=1;(i)<=n;i++)
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
typedef long long ll;
inline void Fail(){printf("0");exit(0);}

const int maxn=200005;
int n,q,ans=INF;
int b[maxn],l[maxn],r[maxn],id[maxn],cnt[maxn][2],dp[maxn];

bool cmp(int a,int b){return l[a]<l[b];}

class segtree{
public:
	int sz;
	int t[maxn*4];
	void _init(int tot){sz=1;for(sz=1;sz<=tot;sz<<=1);rep(i,sz<<1)t[i]=INF;}
	void upd(int pos,int val){
		pos+=sz-1;t[pos]=val;
		for(pos>>=1;pos;pos>>=1)t[pos]=min(t[pos<<1],t[pos<<1|1]);
	}
	int qry(int l,int r){
		int res=INF;
		for(l+=sz-1,r+=sz-1;l<=r;l>>=1,r>>=1){
			if(l&1)res=min(res,t[l]),l++;
			if((r+1)&1)res=min(res,t[r]),r--;
		}
		return res;
	}
}T[2];

int main(){
	scanf("%d",&n);
	REP(i,n){
		scanf("%d",&b[i]);
		rep(j,2)cnt[i][j]=cnt[i-1][j]+(b[i]==j);
	}
	scanf("%d",&q);
	rep(i,q){
		scanf("%d%d",&l[i],&r[i]);
		id[i]=i;
	}
	sort(id,id+q,cmp);
	rep(i,2)T[i]._init(n);
	ans=cnt[n][1];
	rep(i,q){
		int cur=id[i];
//		cout<<cur<<endl;
		dp[cur]=cnt[l[cur]-1][1]+cnt[r[cur]][0]-cnt[l[cur]-1][0];
		dp[cur]=min(dp[cur],T[0].qry(l[cur],r[cur])+cnt[r[cur]][0]);
		dp[cur]=min(dp[cur],T[1].qry(1,l[cur]-1)+cnt[r[cur]][0]-cnt[l[cur]-1][0]+cnt[l[cur]-1][1]);
		T[0].upd(r[cur],dp[cur]-cnt[r[cur]][0]);
		T[1].upd(r[cur],dp[cur]-cnt[r[cur]][1]);
		ans=min(ans,dp[cur]+cnt[n][1]-cnt[r[cur]][1]);
	}
	printf("%d",ans);
	return 0;
}

/*
Input:
-----------------
Output:
*/
### Prompt

Create a solution in cpp for 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
/* --- author: dxm --- */
#include<bits/stdc++.h>
using namespace std;

#define INF 1e9
#define Inf 1e18
#define rep(i,n) for(int (i)=0;(i)<n;i++)
#define REP(i,n) for(int (i)=1;(i)<=n;i++)
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
typedef long long ll;
inline void Fail(){printf("0");exit(0);}

const int maxn=200005;
int n,q,ans=INF;
int b[maxn],l[maxn],r[maxn],id[maxn],cnt[maxn][2],dp[maxn];

bool cmp(int a,int b){return l[a]<l[b];}

class segtree{
public:
	int sz;
	int t[maxn*4];
	void _init(int tot){sz=1;for(sz=1;sz<=tot;sz<<=1);rep(i,sz<<1)t[i]=INF;}
	void upd(int pos,int val){
		pos+=sz-1;t[pos]=val;
		for(pos>>=1;pos;pos>>=1)t[pos]=min(t[pos<<1],t[pos<<1|1]);
	}
	int qry(int l,int r){
		int res=INF;
		for(l+=sz-1,r+=sz-1;l<=r;l>>=1,r>>=1){
			if(l&1)res=min(res,t[l]),l++;
			if((r+1)&1)res=min(res,t[r]),r--;
		}
		return res;
	}
}T[2];

int main(){
	scanf("%d",&n);
	REP(i,n){
		scanf("%d",&b[i]);
		rep(j,2)cnt[i][j]=cnt[i-1][j]+(b[i]==j);
	}
	scanf("%d",&q);
	rep(i,q){
		scanf("%d%d",&l[i],&r[i]);
		id[i]=i;
	}
	sort(id,id+q,cmp);
	rep(i,2)T[i]._init(n);
	ans=cnt[n][1];
	rep(i,q){
		int cur=id[i];
//		cout<<cur<<endl;
		dp[cur]=cnt[l[cur]-1][1]+cnt[r[cur]][0]-cnt[l[cur]-1][0];
		dp[cur]=min(dp[cur],T[0].qry(l[cur],r[cur])+cnt[r[cur]][0]);
		dp[cur]=min(dp[cur],T[1].qry(1,l[cur]-1)+cnt[r[cur]][0]-cnt[l[cur]-1][0]+cnt[l[cur]-1][1]);
		T[0].upd(r[cur],dp[cur]-cnt[r[cur]][0]);
		T[1].upd(r[cur],dp[cur]-cnt[r[cur]][1]);
		ans=min(ans,dp[cur]+cnt[n][1]-cnt[r[cur]][1]);
	}
	printf("%d",ans);
	return 0;
}

/*
Input:
-----------------
Output:
*/
```