#include<bits/stdc++.h>
using namespace std;

#define se second
#define fi first
#define mk(a,b) make_pair(a,b)

typedef pair<int,int> pii;

int N,N_,Q;
int sum[200010];
pii que[200010];
int seg[2][800010];
int INF = 1001001001;
int ans = INF,cnt;

void init()
{
    N_ = 1;
    while(N_ <= N + 1)N_ *= 2;
    for(int i = 0; i < N_ * 2 - 1; i++)seg[0][i] = seg[1][i] = INF;
    return;
}

void update(int id,int k,int x)
{
    k += N_ - 1;
    seg[id][k] = x;
    while(k)
    {
        k = (k - 1) / 2;
        seg[id][k] = min(seg[id][k * 2 + 1],seg[id][k * 2 + 2]);
    }
    return;
}

int get_min(int id,int a,int b,int k = 0,int l = 0,int r = N_)
{
    if(a <= l && r <= b)return seg[id][k];
    if(b <= l || r <= a)return INF;
    int L = get_min(id,a,b,k * 2 + 1,l,(l + r) / 2);
    int R = get_min(id,a,b,k * 2 + 2,(l + r) / 2,r);
    return min(L,R);
}

int main()
{
    ///1index
    scanf("%d",&N);
    for(int i = 1; i <= N; i++)
    {
        int a;
        scanf("%d",&a);
        if(a)cnt++;
        if(a)sum[i] = sum[i - 1] - 1;
        else sum[i] = sum[i - 1] + 1;
    }
    scanf("%d",&Q);
    for(int i = 0; i < Q; i++)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        que[i] = mk(l,r);
    }

    sort(que,que + Q);

    init();
    update(0,0,0);
    update(1,0,0);

    for(int i = 0; i < Q; i++)
    {
        int res = get_min(1,que[i].fi,que[i].se + 1) + sum[que[i].se];
        res = min(res,get_min(0,0,que[i].fi) + sum[que[i].se] - sum[que[i].fi - 1]);
        if(res < get_min(0,que[i].se,que[i].se + 1))
        {
            update(1,que[i].se,res - sum[que[i].se]);
            update(0,que[i].se,res);
        }
    }

    for(int i = 0; i <= N; i++)ans = min(ans,get_min(0,i,i + 1));

    printf("%d\n",ans + cnt);

    return 0;
}

### Prompt

Please formulate 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<bits/stdc++.h>
using namespace std;

#define se second
#define fi first
#define mk(a,b) make_pair(a,b)

typedef pair<int,int> pii;

int N,N_,Q;
int sum[200010];
pii que[200010];
int seg[2][800010];
int INF = 1001001001;
int ans = INF,cnt;

void init()
{
    N_ = 1;
    while(N_ <= N + 1)N_ *= 2;
    for(int i = 0; i < N_ * 2 - 1; i++)seg[0][i] = seg[1][i] = INF;
    return;
}

void update(int id,int k,int x)
{
    k += N_ - 1;
    seg[id][k] = x;
    while(k)
    {
        k = (k - 1) / 2;
        seg[id][k] = min(seg[id][k * 2 + 1],seg[id][k * 2 + 2]);
    }
    return;
}

int get_min(int id,int a,int b,int k = 0,int l = 0,int r = N_)
{
    if(a <= l && r <= b)return seg[id][k];
    if(b <= l || r <= a)return INF;
    int L = get_min(id,a,b,k * 2 + 1,l,(l + r) / 2);
    int R = get_min(id,a,b,k * 2 + 2,(l + r) / 2,r);
    return min(L,R);
}

int main()
{
    ///1index
    scanf("%d",&N);
    for(int i = 1; i <= N; i++)
    {
        int a;
        scanf("%d",&a);
        if(a)cnt++;
        if(a)sum[i] = sum[i - 1] - 1;
        else sum[i] = sum[i - 1] + 1;
    }
    scanf("%d",&Q);
    for(int i = 0; i < Q; i++)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        que[i] = mk(l,r);
    }

    sort(que,que + Q);

    init();
    update(0,0,0);
    update(1,0,0);

    for(int i = 0; i < Q; i++)
    {
        int res = get_min(1,que[i].fi,que[i].se + 1) + sum[que[i].se];
        res = min(res,get_min(0,0,que[i].fi) + sum[que[i].se] - sum[que[i].fi - 1]);
        if(res < get_min(0,que[i].se,que[i].se + 1))
        {
            update(1,que[i].se,res - sum[que[i].se]);
            update(0,que[i].se,res);
        }
    }

    for(int i = 0; i <= N; i++)ans = min(ans,get_min(0,i,i + 1));

    printf("%d\n",ans + cnt);

    return 0;
}

```