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>
#pragma GCC optimize "03"
using namespace std;
const long long int N = 1e5 + 5;
const long long int mod = 1e9 + 7;
const long long int inf = 1e9 + 9;
set<pair<long long int, long long int> > st;
long long int t[4 * N], lz[4 * N], col[N];
void pushdown(long long int nd, long long int s, long long int e) {
  t[nd] += lz[nd] * (e - s + 1);
  if (s != e) {
    lz[nd << 1] += lz[nd];
    lz[nd << 1 | 1] += lz[nd];
  }
  lz[nd] = 0;
}
void upd(long long int nd, long long int s, long long int e, long long int l,
         long long int r, long long int x) {
  pushdown(nd, s, e);
  if (s > e || s > r || e < l) return;
  if (s >= l && e <= r) {
    t[nd] += x * (e - s + 1);
    if (s != e) {
      lz[nd << 1] += x;
      lz[nd << 1 | 1] += x;
    }
    return;
  }
  long long int md = (s + e) >> 1;
  upd(nd << 1, s, md, l, r, x);
  upd(nd << 1 | 1, md + 1, e, l, r, x);
  t[nd] = t[nd << 1] + t[nd << 1 | 1];
}
long long int query(long long int nd, long long int s, long long int e,
                    long long int l, long long int r) {
  pushdown(nd, s, e);
  if (s > e || s > r || e < l) return 0;
  if (s >= l && e <= r) return t[nd];
  long long int md = (s + e) >> 1;
  return query(nd << 1, s, md, l, r) + query(nd << 1 | 1, md + 1, e, l, r);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, m;
  cin >> n >> m;
  for (long long int i = 1; i <= n; i++) {
    st.insert({i, i});
    col[i] = i;
  }
  while (m--) {
    long long int t, l, r, x;
    cin >> t >> l >> r;
    if (t == 1) {
      cin >> x;
      auto p = st.upper_bound({l, inf});
      auto q = p;
      q--;
      if (q->second > r) {
        upd(1, 1, n, l, r, abs(x - col[q->first]));
        st.insert({r + 1, q->second});
        col[r + 1] = col[q->first];
        if (q->first < l) st.insert({q->first, l - 1});
        st.erase(q);
      } else if (q->second >= l) {
        upd(1, 1, n, l, q->second, abs(x - col[q->first]));
        if (q->first < l) st.insert({q->first, l - 1});
        st.erase(q);
      }
      while (!st.empty()) {
        auto p = st.upper_bound({l, inf});
        if (p == st.end() || p->first > r)
          break;
        else if (p->second <= r) {
          upd(1, 1, n, p->first, p->second, abs(x - col[p->first]));
          st.erase(p);
        } else {
          upd(1, 1, n, p->first, r, abs(x - col[p->first]));
          st.erase(p);
          col[r + 1] = col[p->first];
          st.insert({r + 1, p->second});
        }
      }
      col[l] = x;
      st.insert({l, r});
    } else
      cout << query(1, 1, n, l, r) << '\n';
  }
  return 0;
}
