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;
string to_string(const string& str) { return str; }
template <typename T>
string to_string(const set<T>& mys) {
  if (mys.empty()) return "{}";
  string ret = "{";
  for (auto el : mys) ret += to_string(el) + ", ";
  ret.pop_back(), ret.pop_back();
  ret += "}";
  return ret;
}
template <typename T>
string to_string(const pair<T, T>& pr) {
  return "(" + to_string(pr.first) + "," + to_string(pr.second) + ")";
}
template <typename T>
string to_string(const vector<T>& vc, int w) {
  if (vc.empty()) return "";
  if (w + 1 == vc.size()) return to_string(vc[w]);
  return to_string(vc[w]) + "," + to_string(vc, w + 1);
}
template <typename T>
string to_string(const vector<T>& vc) {
  return "{" + to_string(vc, 0) + "}";
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
class DebugStream {
} LOG;
template <typename T>
DebugStream& operator<<(DebugStream& s, const T&) {
  return s;
}
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int maxN = 2e5 + 9, maxV = 1e6 + 9, MOD = 1e9 + 7, SQ = 335, lg = 20,
          bs = 29;
int n;
long long query(long long* T, int l, int r, bool md = true) {
  long long ret = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) ret += T[l++];
    if (r & 1) ret += T[--r];
    if (md) ret %= MOD;
  }
  return ret;
}
void pullUp(long long* T, int w, bool md = true) {
  while (w > 1) {
    w >>= 1;
    T[w] = T[w << 1] + T[w << 1 | 1];
    if (md) T[w] %= MOD;
  }
}
long long pln[maxN << 1];
struct Estr {
  long long arr[maxN], wgt[maxN << 1], cst[maxN << 1];
  void build() {
    for (int i = 0; i < n; i++) cst[i + n] = wgt[i + n] * (arr[i] - i) % MOD;
    for (int i = n - 1; i > 0; i--) {
      cst[i] = (cst[i << 1] + cst[i << 1 | 1]) % MOD;
      wgt[i] = (wgt[i << 1] + wgt[i << 1 | 1]) % MOD;
    }
  }
  void upd(int w, long long x) {
    wgt[w + n] = x;
    cst[w + n] = x * (arr[w] - w) % MOD;
    pullUp(wgt, w + n), pullUp(cst, w + n);
  }
  long long calc(int le, int ri) {
    long long dists = query(cst, le, ri);
    long long sob = query(wgt, le, ri) * (arr[le] - le) % MOD;
    return (dists - sob + MOD) % MOD;
  }
} one, two;
int getKey(int a, int b) {
  int le = a, ri = b;
  while (le + 1 < ri) {
    int mi = (le + ri) / 2;
    if (query(pln, a, mi, false) <= query(pln, mi, b, false))
      le = mi;
    else
      ri = mi;
  }
  return le;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> n >> q;
  for (int i = 0, inp; i < n; i++) {
    cin >> inp;
    one.arr[i] = inp;
    two.arr[n - i - 1] = MOD - inp;
  }
  for (int i = 0, inp; i < n; i++) {
    cin >> inp;
    pln[i + n] = inp;
    one.wgt[i + n] = inp;
    two.wgt[n * 2 - i - 1] = inp;
  }
  for (int i = n - 1; i > 0; i--) pln[i] = pln[i << 1] + pln[i << 1 | 1];
  one.build();
  two.build();
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      int w = -x - 1, nw = y;
      one.upd(w, nw);
      two.upd(n - w - 1, nw);
      pln[w + n] = nw;
      pullUp(pln, w + n, false);
    } else {
      int le = x, ri = y;
      le--;
      int w = getKey(le, ri);
      long long ans = (one.calc(w, ri) + two.calc(n - w - 1, n - le)) % MOD;
      cout << ans << '\n';
    }
  }
}