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;
const double PI = 2 * acos(0.0);
template <typename T>
struct edge {
  int from, to;
  T w;
  edge(int from, int to, T w) : from(from), to(to), w(w) {}
  edge() {}
};
const long double EPS = 1e-9;
template <class T>
void show(T a) {
  for (auto x : a) cout << x << " ";
  cout << endl;
}
const int N = 2e5 + 1;
const int M = 1e9 + 7;
long double pw_double[N], pa_double[N];
long double tw_double[N], ta_double[N];
inline void upd_double(int i, long double w, long double a) {
  long double tmp = w;
  w -= pw_double[i];
  pw_double[i] = tmp;
  tmp = a;
  a -= pa_double[i];
  pa_double[i] = tmp;
  for (++i; i < N; i += i & -i) tw_double[i] += w, ta_double[i] += a;
}
inline void get_double(int l, int r, long double &w, long double &a) {
  if (l > r) return;
  for (++r; r > 0; r -= r & -r) w += tw_double[r], a += ta_double[r];
  for (; l > 0; l -= l & -l) w -= tw_double[l], a -= ta_double[l];
}
int pw[N], pa[N];
int tw[N], ta[N];
inline void upd(int i, int w, int a) {
  int tmp = w;
  w = ((w) - (pw[i]) < 0 ? (w) - (pw[i]) + M : (w) - (pw[i]));
  pw[i] = tmp;
  tmp = a;
  a = ((a) - (pa[i]) < 0 ? (a) - (pa[i]) + M : (a) - (pa[i]));
  pa[i] = tmp;
  for (++i; i < N; i += i & -i)
    tw[i] = ((tw[i]) + (w) >= M ? (tw[i]) + (w)-M : (tw[i]) + (w)),
    ta[i] = ((ta[i]) + (a) >= M ? (ta[i]) + (a)-M : (ta[i]) + (a));
}
inline void get(int l, int r, int &w, int &a) {
  if (l > r) return;
  for (++r; r > 0; r -= r & -r)
    w = ((w) + (tw[r]) >= M ? (w) + (tw[r]) - M : (w) + (tw[r])),
    a = ((a) + (ta[r]) >= M ? (a) + (ta[r]) - M : (a) + (ta[r]));
  for (; l > 0; l -= l & -l)
    w = ((w) - (tw[l]) < 0 ? (w) - (tw[l]) + M : (w) - (tw[l])),
    a = ((a) - (ta[l]) < 0 ? (a) - (ta[l]) + M : (a) - (ta[l]));
}
int a[N];
int n, q;
inline int bins(long long xl) {
  int l = -1, r = n;
  while (r - l > 1) {
    int m = (l + r) >> 1;
    if (xl + m - a[m] >= 0)
      l = m;
    else
      r = m;
  }
  return r;
}
inline long double get_double(int xx, int l, int r) {
  int pos = bins(xx - l);
  long double x = xx;
  long double w = 0, a = 0, w1 = 0, a1 = 0;
  get_double(l, min(pos - 1, r), w, a);
  get_double(max(l, pos), r, w1, a1);
  return (w - w1) * (x - l) + a - a1;
}
inline int get(int x, int l, int r) {
  int pos = bins(x - l);
  int w = 0, a = 0, w1 = 0, a1 = 0;
  get(l, min(pos - 1, r), w, a);
  get(max(l, pos), r, w1, a1);
  return ((((((long long)((w) - (w1) < 0 ? (w) - (w1) + M : (w) - (w1))) *
             (((x) - (l) < 0 ? (x) - (l) + M : (x) - (l)))) %
            M)) + (((a) - (a1) < 0 ? (a) - (a1) + M : (a) - (a1))) >=
                  M
              ? (((((long long)((w) - (w1) < 0 ? (w) - (w1) + M : (w) - (w1))) *
                   (((x) - (l) < 0 ? (x) - (l) + M : (x) - (l)))) %
                  M)) +
                    (((a) - (a1) < 0 ? (a) - (a1) + M : (a) - (a1))) - M
              : (((((long long)((w) - (w1) < 0 ? (w) - (w1) + M : (w) - (w1))) *
                   (((x) - (l) < 0 ? (x) - (l) + M : (x) - (l)))) %
                  M)) +
                    (((a) - (a1) < 0 ? (a) - (a1) + M : (a) - (a1))));
}
inline bool check(int x, int l, int r) {
  long double a = get_double(x, l, r), b = get_double(x + 1, l, r);
  return a > b || abs(a - b) < EPS;
}
int solve(int L, int R) {
  int l = -1, r = 1e9 + 9;
  while (r - l > 1) {
    int m = (l + r) >> 1;
    if (check(m, L, R))
      l = m;
    else
      r = m;
  }
  return get(r, L, R);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < (int)(n); ++i) cin >> a[i];
  int w;
  for (int i = 0; i < (int)(n); ++i) {
    cin >> w;
    upd(i, w,
        ((((long long)w) *
          (((i) - (a[i]) < 0 ? (i) - (a[i]) + M : (i) - (a[i])))) %
         M));
    upd_double(i, w, (long double)w * (i - a[i]));
  }
  int x, y;
  while (q--) {
    cin >> x >> y;
    if (x < 0) {
      x = -x - 1;
      upd(x, y,
          ((((long long)y) *
            (((x) - (a[x]) < 0 ? (x) - (a[x]) + M : (x) - (a[x])))) %
           M));
      upd_double(x, y, (long double)y * (x - a[x]));
    } else {
      cout << solve(--x, --y) << "\n";
    }
  }
}