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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long N;
set<pair<pair<long long, long long>, long long>> st;
struct node {
  long long val, lazy;
  node() { val = lazy = 0; }
};
node tre[3 * 200005];
long long arr[200005];
void propagete(long long n, long long l, long long r) {
  tre[n].val += tre[n].lazy * (r - l + 1);
  if (l != r) {
    tre[2 * n].lazy += tre[n].lazy;
    tre[2 * n + 1].lazy += tre[n].lazy;
  }
  tre[n].lazy = 0;
}
long long combine(long long a, long long b) { return (a + b); }
void update(long long n, long long l, long long r, long long i, long long j,
            long long val) {
  propagete(n, l, r);
  if (l > j || r < i) return;
  if (i <= l && j >= r) {
    tre[n].lazy += val;
    propagete(n, l, r);
    return;
  }
  long long mid = (l + r) / 2;
  update(2 * n, l, mid, i, j, val);
  update(2 * n + 1, mid + 1, r, i, j, val);
  tre[n].val = combine(tre[2 * n].val, tre[2 * n + 1].val);
}
long long query(long long n, long long l, long long r, long long i,
                long long j) {
  propagete(n, l, r);
  if (l > j || r < i) return 0;
  if (i <= l && j >= r) {
    return tre[n].val;
  }
  long long mid = (l + r) / 2;
  return combine(query(2 * n, l, mid, i, j),
                 query(2 * n + 1, mid + 1, r, i, j));
}
void sh(pair<pair<long long, long long>, long long> p, long long l, long long r,
        long long clr) {
  st.erase(p);
  vector<pair<pair<long long, long long>, long long>> v;
  if (p.first.first < l) {
    st.insert({{p.first.first, l - 1}, p.second});
    p.first.first = l;
  }
  if (p.first.second > r) {
    st.insert({{r + 1, p.first.second}, p.second});
    p.first.second = r;
  }
  update(1, 0, N - 1, p.first.first - 1, p.first.second - 1,
         abs(p.second - clr));
  p.second = clr;
  st.insert(p);
}
void modify(long long l, long long r, long long clr) {
  vector<pair<pair<long long, long long>, long long>> v;
  auto cv = st.lower_bound({{l, 0}, 0});
  if (cv != st.begin()) cv--;
  for (cv; cv != st.end(); cv++) {
    if (cv->first.first > r) break;
    if (cv->first.second < l) continue;
    v.push_back(*cv);
  }
  long long x, y, z;
  x = r + 1;
  y = l - 1;
  for (auto &cv : v) {
    if (cv.first.first >= l && cv.first.second <= r) {
      st.erase(cv);
      update(1, 0, N - 1, cv.first.first - 1, cv.first.second - 1,
             abs(cv.second - clr));
      x = min(cv.first.first, x);
      y = max(cv.first.second, y);
      z = abs(clr - cv.second);
    } else
      sh(cv, l, r, clr);
  }
  if (x <= y) {
    st.insert({{x, y}, clr});
  }
}
void solve() {
  long long i, j, k, l, r, m, n, a, b, c, d, x, y, z, t;
  cin >> n >> m;
  N = n;
  for (i = 1; i <= n; i++) {
    st.insert({{i, i}, i});
  }
  for (i = 1; i <= m; i++) {
    cin >> t;
    if (t == 1) {
      cin >> x >> y >> z;
      modify(x, y, z);
    } else {
      cin >> x >> y;
      cout << query(1, 0, N - 1, x - 1, y - 1) << endl;
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}
