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>
#define inf 0x3f3f3f3f
using namespace std;
const int N=2e5+5;
int n,m,sum,a[N],lev[N],t[N<<2];
vector<int>v[N];
void update(int k)
{
    t[k]=min(t[k<<1],t[k<<1|1]);
}
void build(int l,int r,int k)
{
    if(l==r)
    {t[k]=l!=0?inf:0;lev[l]=k;return;}
    int m=l+r>>1;
    build(l,m,k<<1);build(m+1,r,k<<1|1);
    update(k);
}
void fix(int x,int v)
{
    x=lev[x];
    if(v<t[x])
    {
        t[x]=v;
        while(x>>=1) update(x);
    }
}
int query(int l,int r,int k,int x,int y)
{
    if(r<x||l>y) return inf;
    if(l>=x&&r<=y) return t[k];
    int m=l+r>>1;
    return min(query(l,m,k<<1,x,y),query(m+1,r,k<<1|1,x,y));
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]),sum+=a[i]==0;
    scanf("%d",&m);
    for(int i=1;i<=m;i++)
    {
        int l,r;scanf("%d%d",&l,&r);v[l].push_back(r);
    }
    build(0,n,1);
    for(int i=1;i<=n;i++)
    {
        sort(v[i].begin(),v[i].end(),greater<int>());
        for(int j:v[i])
        {
            int ans=query(0,n,1,i-1,j);
            fix(j,ans);
        }
        int ans=query(0,n,1,i-1,i-1);
        ans+=a[i]==0?-1:1;
        fix(i,ans);
    }
    printf("%d\n",query(0,n,1,n,n)+sum);
}