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;
long long int tree[4 * 100005], lazy[4 * 100005], color[4 * 100005];
void propagate(int node, int low, int high) {
  int rng = (high - low + 1);
  color[node] += (long long int)rng * lazy[node];
  if (low != high) {
    lazy[2 * node] += lazy[node];
    lazy[2 * node + 1] += lazy[node];
    if (tree[node] != -1) {
      tree[2 * node] = tree[node];
      tree[2 * node + 1] = tree[node];
    }
  }
  lazy[node] = 0;
}
void go(int node, int low, int high) {
  if (low == high) {
    tree[node] = low;
    return;
  }
  int mid = (low + high) / 2;
  go(2 * node, low, mid);
  go(2 * node + 1, mid + 1, high);
  tree[node] = -1;
}
void upd(int node, int low, int high, int l, int r, long long int x) {
  if (lazy[node] != 0) propagate(node, low, high);
  if (low > r or high < l) return;
  if (low >= l and high <= r and tree[node] != -1) {
    lazy[node] = abs(tree[node] - x);
    tree[node] = x;
    propagate(node, low, high);
    return;
  }
  int mid = (low + high) / 2;
  upd(2 * node, low, mid, l, r, x);
  upd(2 * node + 1, mid + 1, high, l, r, x);
  tree[node] = (tree[2 * node] == tree[2 * node + 1]) ? tree[2 * node] : -1;
  color[node] = color[2 * node] + color[2 * node + 1];
}
long long int ask(int node, int low, int high, int l, int r) {
  if (lazy[node] != 0) propagate(node, low, high);
  if (low > r or high < l) return 0;
  if (low >= l and high <= r) {
    return color[node];
  }
  int mid = (low + high) / 2;
  return ask(2 * node, low, mid, l, r) + ask(2 * node + 1, mid + 1, high, l, r);
}
int main() {
  int n, m;
  cin >> n >> m;
  go(1, 1, n);
  while (m--) {
    int t;
    cin >> t;
    if (t == 1) {
      long long int l, r, x;
      cin >> l >> r >> x;
      upd(1, 1, n, l, r, x);
    } else {
      int l, r;
      cin >> l >> r;
      cout << ask(1, 1, n, l, r) << endl;
    }
  }
  return 0;
}
