DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
#pragma GCC diagnostic ignored "-Wsign-compare"
int n, m;
vector<long long> fa(100001, 0);
vector<long long> fb(100001, 0);
void upd(int left, long long val) {
  long long valupd = val * (n - left);
  for (int i = left + 1; i <= n; i += i & -i) {
    fa[i] += valupd;
    fb[i] += val;
  }
}
long long qry(int right) {
  long long ret = 0;
  long long ret1 = 0;
  long long ret2 = 0;
  for (int i = right + 1; i > 0; i -= i & -i) {
    ret1 += fa[i];
    ret2 += fb[i];
  }
  ret = ret1 - (n - 1 - right) * ret2;
  return ret;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  set<pair<long long, long long>> use;
  for (int i = 0; i <= n; i++) {
    use.insert(make_pair(i, i + 1));
  }
  for (int i = 0; i < m; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      auto it1 = use.upper_bound(make_pair(l + 1, 0));
      auto it2 = use.upper_bound(make_pair(r + 2, 0));
      --it1;
      --it2;
      pair<long long, long long> p1 = *it1;
      pair<long long, long long> p2 = *it2;
      if (p1.first != l) {
        use.insert(make_pair(l, p1.second));
      }
      if (p2.first != r + 1) {
        use.insert(make_pair(r + 1, p2.second));
      }
      it2 = use.lower_bound(make_pair(l, 0));
      p2 = *it2;
      long long val;
      cin >> val;
      while (p2.first <= r) {
        it1 = it2++;
        p1 = *it1;
        p2 = *it2;
        upd(p1.first, abs(val - p1.second));
        upd(p2.first, -abs(val - p1.second));
        it2 = use.erase(it1);
        p2 = *it2;
      }
      use.insert(make_pair(l, val));
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << qry(r) - qry(l - 1) << endl;
    }
  }
  return 0;
}
