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;
ifstream in;
ofstream out;
const long long kk = 1000;
const long long ml = kk * kk;
const long long mod = ml * kk + 7;
const long long inf = ml * ml * ml + 7;
struct triple {
  long long l, r, v;
};
long long lt[ml], rt[ml];
long long t[ml];
long long n, i, j, s, q;
vector<long long> x, w;
vector<triple> seg;
bool viv = false;
void change(long long v, long long tl, long long tr, long long num,
            long long nw) {
  if (tl <= num && num <= tr) {
    t[v] += nw;
    lt[v] += (x[num] - x[tl] - num + tl) * nw;
    long long real_tr = min(tr, n - 1);
    rt[v] += (x[real_tr] - x[num] - real_tr + num) * nw;
    lt[v] %= mod;
    rt[v] %= mod;
    long long tm = tl + tr >> 1;
    if (tl < tr) {
      change(2 * v, tl, tm, num, nw);
      change(2 * v + 1, tm + 1, tr, num, nw);
    }
  }
}
long long getr(long long l, long long r, long long v, long long tl,
               long long tr, long long num) {
  if (r < tl || tr < l) return 0;
  if (l <= tl && tr <= r) {
    long long ret = rt[v];
    long long real_tr = min(tr, n - 1);
    ret += ((t[v] % mod) * (x[num] - x[real_tr] - num + real_tr));
    ret %= mod;
    return ret;
  }
  long long tm = tl + tr >> 1;
  return (getr(l, r, 2 * v, tl, tm, num) +
          getr(l, r, 2 * v + 1, tm + 1, tr, num)) %
         mod;
}
long long getl(long long l, long long r, long long v, long long tl,
               long long tr, long long num) {
  if (r < tl || tr < l) return 0;
  if (l <= tl && tr <= r) {
    long long ret = lt[v];
    ret += ((t[v] % mod) * (x[tl] - x[num] - tl + num));
    ret %= mod;
    return ret;
  }
  long long tm = tl + tr >> 1;
  return (getl(l, r, 2 * v, tl, tm, num) +
          getl(l, r, 2 * v + 1, tm + 1, tr, num)) %
         mod;
}
long long sum(long long l, long long r, long long v, long long tl,
              long long tr) {
  if (r < tl || tr < l) return 0;
  if (l <= tl && tr <= r) return t[v];
  long long tm = tl + tr >> 1;
  return sum(l, r, 2 * v, tl, tm) + sum(l, r, 2 * v + 1, tm + 1, tr);
}
long long found(long long a, long long b) {
  long long l = a, r = b + 1;
  long long locs = sum(a, b, 1, 0, s - 1);
  long long tol = 0;
  while (l + 1 != r) {
    long long mi = l + r >> 1;
    long long ls = tol + sum(l, mi - 1, 1, 0, s - 1);
    if (2 * ls <= locs)
      l = mi, tol = ls;
    else
      r = mi;
  }
  return l;
}
void get(long long l, long long r, long long v, long long tl, long long tr) {
  if (r < tl || tr < l) return;
  if (l <= tl && tr <= r) {
    seg.push_back({tl, tr, t[v]});
    return;
  }
  long long tm = tr + tl >> 1;
  get(l, r, 2 * v, tl, tm);
  get(l, r, 2 * v + 1, tm + 1, tr);
}
long long ver(long long l, long long r, long long v, long long tl,
              long long tr) {
  if (l == tl && r == tr) return v;
  long long tm = tl + tr >> 1;
  if (tm >= r)
    return ver(l, r, 2 * v, tl, tm);
  else
    return ver(l, r, 2 * v + 1, tm + 1, tr);
}
long long gran(long long v, long long l, long long r, long long want) {
  if (l == r) return l;
  if (t[v] == want) return r + 1;
  long long tm = l + r >> 1;
  if (t[2 * v] > want) return gran(2 * v, l, tm, want);
  return gran(2 * v + 1, tm + 1, r, want - t[2 * v]);
}
long long found_new(long long a, long long b) {
  long long locs = sum(a, b, 1, 0, s - 1);
  seg.clear();
  get(a, b, 1, 0, s - 1);
  vector<long long> pr;
  pr.push_back(seg[0].v);
  for (long long i = 1; i < seg.size(); i++) pr.push_back(pr.back() + seg[i].v);
  j = 0;
  while (pr[j] * 2 < locs) j++;
  if (pr[j] * 2 == locs) return seg[j].r + 1;
  long long l = seg[j].l;
  long long r = seg[j].r;
  long long want = locs / 2;
  if (j != 0) want -= pr[j - 1];
  long long v = ver(l, r, 1, 0, s - 1);
  return gran(v, l, r, want);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  x.resize(n);
  w.resize(n);
  for (auto &i : x) cin >> i;
  for (auto &i : w) cin >> i;
  s = 1;
  while (s < n) s *= 2;
  for (i = 0; i < n; i++) change(1, 0, s - 1, i, w[i]);
  if (viv) {
    for (i = 1; i < n + s; i++) cout << t[i] << ' ';
    cout << endl;
    for (i = 1; i < n + s; i++) cout << lt[i] << ' ';
    cout << endl;
    for (i = 1; i < n + s; i++) cout << rt[i] << ' ';
    cout << endl;
  }
  for (i = 0; i < q; i++) {
    long long a, b;
    cin >> a >> b;
    if (a < 0) {
      a = -a;
      a--;
      change(1, 0, s - 1, a, b - w[a]);
      w[a] = b;
    } else {
      a--;
      b--;
      long long mi = found_new(a, b);
      long long lans = 0;
      lans += getr(a, mi - 1, 1, 0, s - 1, mi);
      lans += getl(mi + 1, b, 1, 0, s - 1, mi);
      lans %= mod;
      lans += mod;
      lans %= mod;
      cout << lans << '\n';
    }
  }
  return 0;
}