Problem: There are n officers in the Army of Byteland. Each officer has some power associated with him. The power of the i-th officer is denoted by p_{i}. As the war is fast approaching, the General would like to know the strength of the army.

The strength of an army is calculated in a strange way in Byteland. The General selects a random subset of officers from these n officers and calls this subset a battalion.(All 2^n subsets of the n officers can be chosen equally likely, including empty subset and the subset of all officers).

The strength of a battalion is calculated in the following way:

Let the powers of the chosen officers be a_{1},a_{2},…,a_{k}, where a_1 ≤ a_2 ≤ ... ≤ a_k. The strength of this battalion is equal to a_1a_2 + a_2a_3 + ... + a_{k-1}a_k. (If the size of Battalion is ≤ 1, then the strength of this battalion is 0).

The strength of the army is equal to the expected value of the strength of the battalion.

As the war is really long, the powers of officers may change. Precisely, there will be q changes. Each one of the form i x indicating that p_{i} is changed to x.

You need to find the strength of the army initially and after each of these q updates.

Note that the changes are permanent.

The strength should be found by modulo 10^{9}+7. Formally, let M=10^{9}+7. It can be shown that the answer can be expressed as an irreducible fraction p/q, where p and q are integers and qnot≡ 0 mod M). Output the integer equal to p⋅ q^{-1} mod M. In other words, output such an integer x that 0 ≤ x < M and x ⋅ q ≡ p mod M).

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 3⋅10^{5}) — the number of officers in Byteland's Army.

The second line contains n integers p_{1},p_{2},…,p_{n} (1 ≤ p_{i} ≤ 10^{9}).

The third line contains a single integer q (1 ≤ q ≤ 3⋅10^{5}) — the number of updates.

Each of the next q lines contains two integers i and x (1 ≤ i ≤ n,  1 ≤ x ≤ 10^{9}), indicating that p_{i} is updated to x .

Output

In the first line output the initial strength of the army.

In i-th of the next q lines, output the strength of the army after i-th update.

Examples

Input


2
1 2
2
1 2
2 1


Output


500000004
1
500000004


Input


4
1 2 3 4
4
1 5
2 5
3 5
4 5


Output


625000011
13
62500020
375000027
62500027

Note

In first testcase, initially, there are four possible battalions 

  * {} Strength = 0 
  * {1} Strength = 0 
  * {2} Strength = 0 
  * {1,2} Strength = 2 

So strength of army is (0+0+0+2)/(4) = 1/2

After changing p_{1} to 2, strength of battallion {1,2} changes to 4, so strength of army becomes 1.

After changing p_{2} to 1, strength of battalion {1,2} again becomes 2, so strength of army becomes 1/2.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int inf = 0x3f3f3f3f, Inf = 0x7fffffff;
const long long INF = 0x7fffffffffffffff;
const double eps = 1e-10;
unsigned int seed = 19260817;
const unsigned int _RAND_MAX_ = 4294967295u;
__inline__ __attribute__((always_inline)) unsigned int Rand() {
  return seed = seed * 998244353u + 1000000007u;
}
template <typename _Tp>
_Tp gcd(const _Tp &a, const _Tp &b) {
  return (!b) ? a : gcd(b, a % b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp abs(const _Tp &a) {
  return a >= 0 ? a : -a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp max(const _Tp &a, const _Tp &b) {
  return a < b ? b : a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp min(const _Tp &a, const _Tp &b) {
  return a < b ? a : b;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmax(_Tp &a, const _Tp &b) {
  (a < b) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmin(_Tp &a, const _Tp &b) {
  (b < a) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) bool _cmp(const _Tp &a,
                                                    const _Tp &b) {
  return abs(a - b) <= eps;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void read(_Tp &x) {
  register char ch(getchar());
  bool f(false);
  while (ch < 48 || ch > 57) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (ch >= 48 && ch <= 57)
    x = (((x << 2) + x) << 1) + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
__inline__ __attribute__((always_inline)) void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
__inline__ __attribute__((always_inline)) int read_str(char *s) {
  register char ch(getchar());
  while (ch == ' ' || ch == '\r' || ch == '\n') ch = getchar();
  register char *tar = s;
  *tar = ch, ch = getchar();
  while (ch != ' ' && ch != '\r' && ch != '\n' && ch != EOF)
    *(++tar) = ch, ch = getchar();
  return tar - s + 1;
}
const int N = 300005;
const int mod = 1000000007;
const long long inv2 = 500000004;
struct matrix {
  long long a, b, c, d, e, f;
  __inline__ __attribute__((always_inline)) matrix operator*(
      const matrix &o) const {
    return (matrix){a * o.a % mod,
                    (b * o.a + c * o.b) % mod,
                    c * o.c % mod,
                    (d * o.a + e * o.b + f * o.d) % mod,
                    (e * o.c + f * o.e) % mod,
                    f * o.f % mod};
  }
} I;
__inline__ __attribute__((always_inline)) matrix get1(long long x) {
  return (matrix){1ll, x * inv2 % mod, inv2, 0, x * inv2 % mod, 1ll};
}
__inline__ __attribute__((always_inline)) matrix get2(long long x) {
  return (matrix){1ll, mod - x, 2ll, (x * x) % mod * inv2 % mod, mod - x, 1ll};
}
int t[N << 1];
int a[N];
int b[N << 1];
struct seg_tr {
  struct Node {
    int ls, rs;
    matrix val;
  } f[N << 2];
  int node_cnt;
  __inline__ __attribute__((always_inline)) void PushUp(int x) {
    f[x].val = f[f[x].ls].val * f[f[x].rs].val;
  }
  int build(int l, int r) {
    int cur = ++node_cnt;
    if (l == r) {
      matrix res(I), a = get1(t[l]);
      while (b[l]) res = res * a, --b[l];
      f[cur].val = res;
      return cur;
    }
    int mid = (l + r) >> 1;
    f[cur].ls = build(l, mid);
    f[cur].rs = build(mid + 1, r);
    PushUp(cur);
    return cur;
  }
  void Update1(int l, int r, int pos, int cur) {
    if (l == r) {
      f[cur].val = f[cur].val * get1(t[l]);
      return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid)
      Update1(l, mid, pos, f[cur].ls);
    else
      Update1(mid + 1, r, pos, f[cur].rs);
    PushUp(cur);
  }
  void Update2(int l, int r, int pos, int cur) {
    if (l == r) {
      f[cur].val = f[cur].val * get2(t[l]);
      return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid)
      Update2(l, mid, pos, f[cur].ls);
    else
      Update2(mid + 1, r, pos, f[cur].rs);
    PushUp(cur);
  }
  __inline__ __attribute__((always_inline)) long long Query() {
    return f[1].val.d;
  }
} tr;
int Q[N][2];
int main() {
  I = (matrix){1ll, 0ll, 1ll, 0ll, 0ll, 1ll};
  int n;
  read(n);
  int pos = 0;
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    t[++pos] = a[i];
  }
  int q;
  read(q);
  for (int i = 1; i <= q; ++i) {
    read(Q[i][0], Q[i][1]);
    t[++pos] = Q[i][1];
  }
  std::sort(t + 1, t + pos + 1);
  pos = std::unique(t + 1, t + pos + 1) - t - 1;
  for (int i = 1; i <= n; ++i) {
    a[i] = std::lower_bound(t + 1, t + pos + 1, a[i]) - t;
    ++b[a[i]];
  }
  tr.build(1, pos);
  printf("%lld\n", tr.Query());
  for (int i = 1; i <= q; ++i) {
    Q[i][1] = std::lower_bound(t + 1, t + pos + 1, Q[i][1]) - t;
    tr.Update2(1, pos, a[Q[i][0]], 1);
    tr.Update1(1, pos, Q[i][1], 1);
    a[Q[i][0]] = Q[i][1];
    printf("%lld\n", tr.Query());
  }
  return 0;
}