Problem: There is an infinite line consisting of cells. There are n boxes in some cells of this line. The i-th box stands in the cell a_i and has weight w_i. All a_i are distinct, moreover, a_{i - 1} < a_i holds for all valid i.

You would like to put together some boxes. Putting together boxes with indices in the segment [l, r] means that you will move some of them in such a way that their positions will form some segment [x, x + (r - l)].

In one step you can move any box to a neighboring cell if it isn't occupied by another box (i.e. you can choose i and change a_i by 1, all positions should remain distinct). You spend w_i units of energy moving the box i by one cell. You can move any box any number of times, in arbitrary order.

Sometimes weights of some boxes change, so you have queries of two types: 

  1. id nw — weight w_{id} of the box id becomes nw. 
  2. l r — you should compute the minimum total energy needed to put together boxes with indices in [l, r]. Since the answer can be rather big, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7. Note that the boxes are not moved during the query, you only should compute the answer. 



Note that you should minimize the answer, not its remainder modulo 10^9 + 7. So if you have two possible answers 2 ⋅ 10^9 + 13 and 2 ⋅ 10^9 + 14, you should choose the first one and print 10^9 + 6, even though the remainder of the second answer is 0.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the number of boxes and the number of queries.

The second line contains n integers a_1, a_2, ... a_n (1 ≤ a_i ≤ 10^9) — the positions of the boxes. All a_i are distinct, a_{i - 1} < a_i holds for all valid i.

The third line contains n integers w_1, w_2, ... w_n (1 ≤ w_i ≤ 10^9) — the initial weights of the boxes.

Next q lines describe queries, one query per line.

Each query is described in a single line, containing two integers x and y. If x < 0, then this query is of the first type, where id = -x, nw = y (1 ≤ id ≤ n, 1 ≤ nw ≤ 10^9). If x > 0, then the query is of the second type, where l = x and r = y (1 ≤ l_j ≤ r_j ≤ n). x can not be equal to 0.

Output

For each query of the second type print the answer on a separate line. Since answer can be large, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7.

Example

Input

5 8
1 2 6 7 10
1 1 1 1 2
1 1
1 5
1 3
3 5
-3 5
-1 10
1 4
2 5


Output

0
10
3
4
18
7

Note

Let's go through queries of the example: 

  1. 1\ 1 — there is only one box so we don't need to move anything. 
  2. 1\ 5 — we can move boxes to segment [4, 8]: 1 ⋅ |1 - 4| + 1 ⋅ |2 - 5| + 1 ⋅ |6 - 6| + 1 ⋅ |7 - 7| + 2 ⋅ |10 - 8| = 10. 
  3. 1\ 3 — we can move boxes to segment [1, 3]. 
  4. 3\ 5 — we can move boxes to segment [7, 9]. 
  5. -3\ 5 — w_3 is changed from 1 to 5. 
  6. -1\ 10 — w_1 is changed from 1 to 10. The weights are now equal to w = [10, 1, 5, 1, 2]. 
  7. 1\ 4 — we can move boxes to segment [1, 4]. 
  8. 2\ 5 — we can move boxes to segment [5, 8]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast", "unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long inf = numeric_limits<long long>::max() / 2;
const long double eps = 1e-9;
const long double pi = acos(-1);
inline void solve(), read();
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read();
  solve();
  return 0;
}
const long long R = 1e9 + 7;
struct fenwick {
  long long n;
  vector<long long> t;
  long long r = 0;
  fenwick() {}
  fenwick(long long n_) {
    n = n_;
    t.resize(n);
  }
  long long f(long long i) { return i & (i + 1); }
  long long h(long long i) { return i | (i + 1); }
  void inc(long long ind, long long d) {
    if (r) {
      d %= R;
      d += R;
      d %= R;
    }
    for (long long i = ind; i < n; i = h(i)) {
      t[i] += d;
      if (r) {
        t[i] %= R;
      }
    }
  }
  long long sum(long long ind) {
    long long ans = 0;
    for (long long i = ind; i >= 0; i = f(i) - 1) {
      ans += t[i];
      if (r) {
        ans %= R;
      }
    }
    return ans;
  }
  long long sum(long long l, long long r) {
    if (!this->r) {
      return sum(r) - sum(l - 1);
    } else {
      return (sum(r) - sum(l - 1) + R) % R;
    }
  }
};
long long n, q;
vector<long long> a, w;
fenwick f, g;
long long binsearch(long long L, long long R) {
  long long l = L, r = R;
  long long s = f.sum(l, r);
  while (r - l > 1) {
    long long m = (l + r) / 2;
    if (2 * f.sum(L, m) > s) {
      r = m;
    } else {
      l = m;
    }
  }
  if (2 * f.sum(L, l) >= s) {
    return l;
  } else {
    return r;
  }
}
inline void query1(long long x, long long w) {
  f.inc(x, -f.sum(x, x));
  f.inc(x, w);
  g.inc(x, -g.sum(x, x));
  g.inc(x, a[x] * w);
}
inline long long query2(long long l, long long r) {
  long long k = binsearch(l, r);
  long long ansl =
      (f.sum(l, k - 1) % R * a[k] % R - g.sum(l, k - 1) % R + R) % R;
  long long ansr =
      (g.sum(k + 1, r) % R - f.sum(k + 1, r) % R * a[k] % R + R) % R;
  long long ans = (ansl + ansr) % R;
  return ans;
}
inline void solve() {
  f = fenwick(n + 1), g = fenwick(n + 1);
  g.r = 1;
  for (long long i = 1; i <= n; i++) {
    f.inc(i, w[i]);
    g.inc(i, w[i] * a[i]);
  }
  for (long long i = 0; i < q; i++) {
    long long x, y;
    cin >> x >> y;
    if (x < 0) {
      query1(-x, y);
    } else {
      long long ans = query2(x, y);
      cout << ans << "\n";
    }
  }
}
inline void read() {
  cin >> n >> q;
  a.resize(n + 1), w.resize(n + 1);
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] -= i;
  }
  for (long long i = 1; i <= n; i++) {
    cin >> w[i];
  }
}