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 long long MAXN = 1e5 + 10;
struct node {
  long long val, diff, lazy, max_val, min_val;
  node() {
    val = 0;
    diff = 0;
    lazy = -1;
    max_val = 0;
    min_val = 0;
  }
};
node segt[4 * MAXN];
node merge_node(node x, node y) {
  node res;
  res.val = x.val + y.val;
  res.max_val = max(x.max_val, y.max_val);
  res.min_val = min(x.min_val, y.min_val);
  return res;
}
void lzu(long long id, long long L, long long R) {
  if (L != R && segt[id].lazy != -1) {
    long long mid = (L + R) / 2;
    segt[2 * id + 1].val += segt[id].diff * (mid - L + 1);
    segt[2 * id + 2].val += segt[id].diff * (R - mid);
    segt[2 * id + 1].max_val = segt[id].lazy;
    segt[2 * id + 2].max_val = segt[id].lazy;
    segt[2 * id + 1].min_val = segt[id].lazy;
    segt[2 * id + 2].min_val = segt[id].lazy;
    segt[2 * id + 1].lazy = segt[id].lazy;
    segt[2 * id + 2].lazy = segt[id].lazy;
    segt[2 * id + 1].diff += segt[id].diff;
    segt[2 * id + 2].diff += segt[id].diff;
  }
  segt[id].diff = 0;
  segt[id].lazy = -1;
}
void dfs(long long id, long long L, long long R, long long x, long long t) {
  lzu(id, L, R);
  if (segt[id].max_val == segt[id].min_val) {
    segt[id].val += (R - L + 1) * abs(x - segt[id].max_val) * t;
    segt[id].diff += abs(x - segt[id].max_val) * t;
    segt[id].max_val = x;
    segt[id].min_val = x;
    segt[id].lazy = x;
    return;
  }
  long long mid = (L + R) / 2;
  dfs(2 * id + 1, L, mid, x, t);
  dfs(2 * id + 2, mid + 1, R, x, t);
  segt[id] = merge_node(segt[2 * id + 1], segt[2 * id + 2]);
}
void update(long long id, long long L, long long R, long long l, long long r,
            long long x, long long t) {
  if (L > r || R < l) {
    return;
  }
  if (L >= l && R <= r) {
    dfs(id, L, R, x, t);
    return;
  }
  lzu(id, L, R);
  long long mid = (L + R) / 2;
  update(2 * id + 1, L, mid, l, r, x, t);
  update(2 * id + 2, mid + 1, R, l, r, x, t);
  segt[id] = merge_node(segt[2 * id + 1], segt[2 * id + 2]);
}
long long get_sum(long long id, long long L, long long R, long long l,
                  long long r) {
  if (L > r || R < l) {
    return 0;
  }
  if (L >= l && R <= r) {
    return segt[id].val;
  }
  lzu(id, L, R);
  long long mid = (L + R) / 2;
  return get_sum(2 * id + 1, L, mid, l, r) +
         get_sum(2 * id + 2, mid + 1, R, l, r);
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  long long n, q;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    update(0, 1, n, i, i, i, 0);
  }
  for (long long i = 0; i < q; i++) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long l, r, x;
      cin >> l >> r >> x;
      update(0, 1, n, l, r, x, 1);
    }
    if (t == 2) {
      long long l, r;
      cin >> l >> r;
      cout << get_sum(0, 1, n, l, r) << endl;
    }
  }
}
