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>
using namespace std;
int n, q;
int nn;
int *arra, *arrw;
long long int *suml, *sumr, *sum;
long long int mod = 1E9 + 7;
int mx;
long sm;
bool TEST = 0;
void read() {
  cin >> n >> q;
  nn = ceil(log2(n)) * n - 1;
  if (nn < 5) nn = 5;
  arra = new int[n];
  arrw = new int[n];
  suml = new long long int[nn];
  sumr = new long long int[nn];
  sum = new long long int[nn];
  sm = 0;
  for (int i = 0; i < n; i++) {
    cin >> arra[i];
    arra[i]--;
  }
  for (int i = 0; i < n; i++) {
    cin >> arrw[i];
    sm += arrw[i];
  }
  mx = arra[n - 1];
}
long long int getValLeft(int ind) {
  long long int w = arrw[ind];
  w *= arra[ind] - ((long long int)ind);
  w %= mod;
  return w;
}
long long int getVal(int ind) { return arrw[ind]; }
long long int getValRight(int ind) {
  long long int w = arrw[ind];
  w *= ((long long int)mx) - ((long long int)arra[ind]) -
       ((long long int)(n - ind - 1));
  w %= mod;
  return w;
}
long long int buildSeg(long long int *arr, long long int (*val)(int), int s,
                       int e, int i, bool ismod = true) {
  if (s == e) {
    arr[i] = val(s);
    if (ismod) arr[i] %= mod;
    return arr[i];
  }
  int m = (s + e) / 2;
  arr[i] = (buildSeg(arr, val, s, m, i * 2 + 1, ismod) +
            buildSeg(arr, val, m + 1, e, i * 2 + 2, ismod));
  if (ismod) arr[i] %= mod;
  return arr[i];
}
long long int sumSeg(long long int *arr, int s, int e, int i, int rs, int re,
                     bool ismod = true) {
  if (s >= rs && e <= re) return arr[i];
  if (e < rs || s > re) return 0;
  int m = (s + e) / 2;
  long long int res = sumSeg(arr, s, m, i * 2 + 1, rs, re, ismod);
  res += sumSeg(arr, m + 1, e, i * 2 + 2, rs, re, ismod);
  if (ismod) res %= mod;
  return res;
}
void updateSeg(long long int *arr, int s, int e, int i, int ind,
               long long int diff, bool ismod = true) {
  if (ind < s || ind > e) return;
  arr[i] = (arr[i] + diff);
  if (ismod) arr[i] %= mod;
  if (s == e) return;
  int m = (s + e) / 2;
  updateSeg(arr, s, m, i * 2 + 1, ind, diff, ismod);
  updateSeg(arr, m + 1, e, i * 2 + 2, ind, diff, ismod);
}
int segIndex(long long int *arr, int s, int e, int i, long long int sml) {
  if (s == e) return s;
  int m = (s + e) / 2;
  if (arr[i * 2 + 1] > sml) return segIndex(arr, s, m, i * 2 + 1, sml);
  return segIndex(arr, m + 1, e, i * 2 + 2, sml - arr[i * 2 + 1]);
}
int getMedian(int s, int e) {
  long long int sml = sumSeg(sum, 0, n - 1, 0, s, e, false);
  long long int smlb = 0;
  if (s > 0) smlb = sumSeg(sum, 0, n - 1, 0, 0, s - 1, false);
  int ind = segIndex(sum, 0, n - 1, 0, smlb + sml / 2);
  return ind;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  buildSeg(suml, getValLeft, 0, n - 1, 0);
  buildSeg(sumr, getValRight, 0, n - 1, 0);
  buildSeg(sum, getVal, 0, n - 1, 0, false);
  if (TEST) {
    cout << endl << "--------------" << endl;
    for (int i = 0; i < nn; i++) cout << sum[i] << " ";
    cout << endl;
    for (int i = 0; i < n; i++) cout << getVal(i) << " ";
    cout << endl;
    for (int i = 0; i < n; i++) cout << getValLeft(i) << " ";
    cout << endl;
    for (int i = 0; i < n; i++) cout << getValRight(i) << " ";
    cout << endl;
    cout << "--------------" << endl;
  }
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      x = -x;
      x--;
      int vl = getValLeft(x);
      int vr = getValRight(x);
      int vm = getVal(x);
      arrw[x] = y;
      int nl = getValLeft(x);
      int nr = getValRight(x);
      int nm = y;
      int dl = ((nl - vl) + mod * 2) % mod;
      int dr = ((nr - vr) + mod * 2) % mod;
      int dm = nm - vm;
      updateSeg(suml, 0, n - 1, 0, x, dl);
      updateSeg(sumr, 0, n - 1, 0, x, dr);
      updateSeg(sum, 0, n - 1, 0, x, dm, false);
    } else {
      x--;
      y--;
      int med = getMedian(x, y);
      long long int need = 0;
      int pb, pa, tb, ta;
      pa = med + 1;
      pb = med;
      long long int smm = sumSeg(sum, 0, n - 1, 0, x, y, false);
      if (TEST) {
        cout << "-------------" << endl;
        cout << smm << " " << med << endl;
      }
      if (smm % 2) {
        pb--;
        tb = arra[med] - 1;
        ta = arra[med] + 1;
      } else {
        long long int sb = 0;
        if (med > x) sb = sumSeg(sum, 0, n - 1, 0, x, med - 1, false);
        if (sb == smm / 2) {
          pa = pb;
          pb = pb - 1;
          tb = (arra[pb] + arra[pa]) / 2;
          ta = tb + 1;
        } else {
          pb--;
          tb = arra[med] - 1;
          ta = arra[med] + 1;
        }
      }
      if (TEST) {
        cout << pb << " " << pa << endl;
        cout << tb << " " << ta << endl;
      }
      if (pb >= x) {
        if (TEST)
          cout << "L: " << (sumSeg(sumr, 0, n - 1, 0, x, pb) % mod) << " - "
               << (sumSeg(sum, 0, n - 1, 0, x, pb)) << " * "
               << (((mx - tb - (n - pb)) + 1) % mod) << endl;
        need = ((need + sumSeg(sumr, 0, n - 1, 0, x, pb) % mod -
                 (sumSeg(sum, 0, n - 1, 0, x, pb) *
                  (((mx - tb - (n - pb)) + 1) % mod)) %
                     mod) +
                mod) %
               mod;
      }
      if (pa <= y) {
        if (TEST)
          cout << "R: " << (sumSeg(suml, 0, n - 1, 0, pa, y) % mod) << " - "
               << (sumSeg(sum, 0, n - 1, 0, pa, y)) << " * " << (ta - pa)
               << endl;
        need = ((need + sumSeg(suml, 0, n - 1, 0, pa, y) % mod -
                 (sumSeg(sum, 0, n - 1, 0, pa, y) * (ta - pa)) % mod) +
                mod) %
               mod;
      }
      if (TEST) cout << "-------------" << endl;
      cout << need << endl;
    }
  }
  return 0;
}