#include <stdio.h>
#include <algorithm>
using namespace std;
struct segment_tree
{
    int seg[524288];
    segment_tree()
    {
        for (int i=0;i<262144;++i)
            seg[262144+i]=i?1000000000:0;
        for (int i=262144-1;i;--i)
            seg[i]=min(seg[i<<1],seg[(i<<1)^1]);
    }
    int query(int id,int il,int ir,int l,int r)
    {
        if (il>=l&&ir<=r)
            return seg[id];
        if (r<=il||l>=ir)
            return 1000000000;
        return min(query(id<<1,il,(il+ir)>>1,l,r),query((id<<1)^1,(il+ir)>>1,ir,l,r));
    }
    void update(int p,int v)
    {
        if (v>=seg[262144+p])
            return;
        seg[262144+p]=v;
        for (int id=(262144+p)>>1;id;id>>=1)
            seg[id]=min(seg[id<<1],seg[(id<<1)^1]);
    }
};
segment_tree orig,added;
int p[200001];
pair<int,int> q[200001];
bool cmp(pair<int,int> a,pair<int,int> b)
{
    return a.second<b.second;
}
int main()
{
    int n;
    scanf("%d",&n);
    int num1=0;
    for (int i=1;i<=n;++i)
    {
        int x;
        scanf("%d",&x);
        p[i]=p[i-1]+(x==0?1:-1);
        num1+=x;
    }
    int nq;
    scanf("%d",&nq);
    for (int i=1;i<=nq;++i)
        scanf("%d%d",&q[i].first,&q[i].second);
    sort(q+1,q+nq+1);
    int mindelta=0;
    for (int i=1;i<=nq;++i)
    {
        int s=q[i].first,e=q[i].second;
        int curv=min(orig.query(1,0,262144,0,s)+p[e]-p[s-1],added.query(1,0,262144,s,e+1)+p[e]);
        if (curv<mindelta)
            mindelta=curv;
        orig.update(e,curv);
        added.update(e,curv-p[e]);
    }
    printf("%d\n",mindelta+num1);
    return 0;
}

### Prompt

In CPP, your task is to solve 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 <stdio.h>
#include <algorithm>
using namespace std;
struct segment_tree
{
    int seg[524288];
    segment_tree()
    {
        for (int i=0;i<262144;++i)
            seg[262144+i]=i?1000000000:0;
        for (int i=262144-1;i;--i)
            seg[i]=min(seg[i<<1],seg[(i<<1)^1]);
    }
    int query(int id,int il,int ir,int l,int r)
    {
        if (il>=l&&ir<=r)
            return seg[id];
        if (r<=il||l>=ir)
            return 1000000000;
        return min(query(id<<1,il,(il+ir)>>1,l,r),query((id<<1)^1,(il+ir)>>1,ir,l,r));
    }
    void update(int p,int v)
    {
        if (v>=seg[262144+p])
            return;
        seg[262144+p]=v;
        for (int id=(262144+p)>>1;id;id>>=1)
            seg[id]=min(seg[id<<1],seg[(id<<1)^1]);
    }
};
segment_tree orig,added;
int p[200001];
pair<int,int> q[200001];
bool cmp(pair<int,int> a,pair<int,int> b)
{
    return a.second<b.second;
}
int main()
{
    int n;
    scanf("%d",&n);
    int num1=0;
    for (int i=1;i<=n;++i)
    {
        int x;
        scanf("%d",&x);
        p[i]=p[i-1]+(x==0?1:-1);
        num1+=x;
    }
    int nq;
    scanf("%d",&nq);
    for (int i=1;i<=nq;++i)
        scanf("%d%d",&q[i].first,&q[i].second);
    sort(q+1,q+nq+1);
    int mindelta=0;
    for (int i=1;i<=nq;++i)
    {
        int s=q[i].first,e=q[i].second;
        int curv=min(orig.query(1,0,262144,0,s)+p[e]-p[s-1],added.query(1,0,262144,s,e+1)+p[e]);
        if (curv<mindelta)
            mindelta=curv;
        orig.update(e,curv);
        added.update(e,curv-p[e]);
    }
    printf("%d\n",mindelta+num1);
    return 0;
}

```